• + ACCUPLACER Mathematics
  • + ACT Mathematics
  • + AFOQT Mathematics
  • + ALEKS Tests
  • + ASVAB Mathematics
  • + ATI TEAS Math Tests
  • + Common Core Math
  • + DAT Math Tests
  • + FSA Tests
  • + FTCE Math
  • + GED Mathematics
  • + Georgia Milestones Assessment
  • + GRE Quantitative Reasoning
  • + HiSET Math Exam
  • + HSPT Math
  • + ISEE Mathematics
  • + PARCC Tests
  • + Praxis Math
  • + PSAT Math Tests
  • + PSSA Tests
  • + SAT Math Tests
  • + SBAC Tests
  • + SIFT Math
  • + SSAT Math Tests
  • + STAAR Tests
  • + TABE Tests
  • + TASC Math
  • + TSI Mathematics
  • + ACT Math Worksheets
  • + Accuplacer Math Worksheets
  • + AFOQT Math Worksheets
  • + ALEKS Math Worksheets
  • + ASVAB Math Worksheets
  • + ATI TEAS 6 Math Worksheets
  • + FTCE General Math Worksheets
  • + GED Math Worksheets
  • + 3rd Grade Mathematics Worksheets
  • + 4th Grade Mathematics Worksheets
  • + 5th Grade Mathematics Worksheets
  • + 6th Grade Math Worksheets
  • + 7th Grade Mathematics Worksheets
  • + 8th Grade Mathematics Worksheets
  • + 9th Grade Math Worksheets
  • + HiSET Math Worksheets
  • + HSPT Math Worksheets
  • + ISEE Middle-Level Math Worksheets
  • + PERT Math Worksheets
  • + Praxis Math Worksheets
  • + PSAT Math Worksheets
  • + SAT Math Worksheets
  • + SIFT Math Worksheets
  • + SSAT Middle Level Math Worksheets
  • + 7th Grade STAAR Math Worksheets
  • + 8th Grade STAAR Math Worksheets
  • + THEA Math Worksheets
  • + TABE Math Worksheets
  • + TASC Math Worksheets
  • + TSI Math Worksheets
  • + AFOQT Math Course
  • + ALEKS Math Course
  • + ASVAB Math Course
  • + ATI TEAS 6 Math Course
  • + CHSPE Math Course
  • + FTCE General Knowledge Course
  • + GED Math Course
  • + HiSET Math Course
  • + HSPT Math Course
  • + ISEE Upper Level Math Course
  • + SHSAT Math Course
  • + SSAT Upper-Level Math Course
  • + PERT Math Course
  • + Praxis Core Math Course
  • + SIFT Math Course
  • + 8th Grade STAAR Math Course
  • + TABE Math Course
  • + TASC Math Course
  • + TSI Math Course
  • + Number Properties Puzzles
  • + Algebra Puzzles
  • + Geometry Puzzles
  • + Intelligent Math Puzzles
  • + Ratio, Proportion & Percentages Puzzles
  • + Other Math Puzzles

How to Solve Permutations and Combinations? (+FREE Worksheet!)

Learn how to solve mathematics word problems containing Permutations and Combinations using formulas.

How to Solve Permutations and Combinations? (+FREE Worksheet!)

Related Topics

  • How to Interpret Histogram
  • How to Interpret Pie Graphs
  • How to Solve Probability Problems
  • How to Find Mean, Median, Mode, and Range of the Given Data

Step by step guide to solve Permutations and Combinations

  • Permutations: The number of ways to choose a sample of \(k\) elements from a set of \(n\) distinct objects where order does matter, and replacements are not allowed. For a permutation problem, use this formula: \(\color{blue}{_{n}P_{k }= \frac{n!}{(n-k)!}}\)
  • Combination: The number of ways to choose a sample of \(r\) elements from a set of \(n\) distinct objects where order does not matter, and replacements are not allowed. For a combination problem, use this formula: \(\color{blue}{_{ n}C_{r }= \frac{n!}{r! (n-r)!}}\)
  • Factorials are products, indicated by an exclamation mark. For example, \(4!\) Equals: \(4×3×2×1\). Remember that \(0!\) is defined to be equal to \(1\).

The Absolute Best Books to Ace Pre-Algebra to Algebra II

The Ultimate Algebra Bundle From Pre-Algebra to Algebra II

Permutations and combinations – example 1:.

How many ways can the first and second place be awarded to \(10\) people?

Since the order matters, we need to use the permutation formula where \(n\) is \(10\) and \(k\) is \(2\). Then: \(\frac{n!}{(n-k)!}=\frac{10!}{(10-2)!}=\frac{10!}{8!}=\frac{10×9×8!}{8!}\), remove \(8!\) from both sides of the fraction. Then: \(\frac{10×9×8!}{8!}=10×9=90\)

Permutations and Combinations – Example 2:

How many ways can we pick a team of \(3\) people from a group of \(8\)?

Since the order doesn’t matter, we need to use a combination formula where \(n\) is \(8\) and \(r\) is \(3\). Then: \(\frac{n!}{r! (n-r)!}=\frac{8!}{3! (8-3)!}=\frac{8!}{3! (5)!}=\frac{8×7×6×5!}{3! (5)!}\), remove \(5!\) from both sides of the fraction. Then: \(\frac{8×7×6}{3×2×1}=\frac{336}{6}=56\)

Exercises for Solving Permutations and Combinations

The Best Book to Help You Ace Pre-Algebra

Pre-Algebra for Beginners The Ultimate Step by Step Guide to Preparing for the Pre-Algebra Test

Calculate the value of each..

  • \(\color{blue}{4!=}\)
  • \(\color{blue}{4!×3!=}\)
  • \(\color{blue}{5!=}\)
  • \(\color{blue}{6!+3!=}\)
  • There are \(7\) horses in a race. In how many different orders can the horses finish?
  • In how many ways can \(6\) people be arranged in a row?

Download Combinations and Permutations Worksheet

  • \(\color{blue}{24}\)
  • \(\color{blue}{144}\)
  • \(\color{blue}{120}\)
  • \(\color{blue}{726}\)
  • \(\color{blue}{5,040}\)
  • \(\color{blue}{720}\)

The Greatest Books for Students to Ace the Algebra

Pre-Algebra Exercise Book A Comprehensive Workbook + PreAlgebra Practice Tests

Pre-algebra in 10 days the most effective pre-algebra crash course, college algebra practice workbook the most comprehensive review of college algebra, high school algebra i a comprehensive review and step-by-step guide to mastering high school algebra 1, 10 full length clep college algebra practice tests the practice you need to ace the clep college algebra test.

by: Effortless Math Team about 4 years ago (category: Articles , Free Math Worksheets )

Effortless Math Team

Related to this article, more math articles.

  • 6th Grade Georgia Milestones Assessment System math Practice Test Questions
  • What is a Good PSAT Score?
  • How to Use Benchmark Fraction? Benchmark Fraction Definition
  • How to Find the y-Intercept of a Line?
  • 8th Grade MEAP Math Practice Test Questions
  • Equivalent Rates
  • How to Convey Decimals in Words
  • The Ultimate PEAKS Algebra 1 Course (+FREE Worksheets)
  • How to Solve Word Problems of Subtracting Numbers Up to 7 Digits
  • AFOQT Math Practice Test Questions

What people say about "How to Solve Permutations and Combinations? (+FREE Worksheet!) - Effortless Math: We Help Students Learn to LOVE Mathematics"?

No one replied yet.

Leave a Reply Cancel reply

You must be logged in to post a comment.

Pre-Algebra Practice Workbook The Most Comprehensive Review of Pre-Algebra

Algebra i practice workbook the most comprehensive review of algebra 1, algebra ii practice workbook the most comprehensive review of algebra 2, algebra i for beginners the ultimate step by step guide to acing algebra i, algebra ii for beginners the ultimate step by step guide to acing algebra ii, pre-algebra tutor everything you need to help achieve an excellent score.

  • ATI TEAS 6 Math
  • ISEE Upper Level Math
  • SSAT Upper-Level Math
  • Praxis Core Math
  • 8th Grade STAAR Math

Limited time only!

Save Over 45 %

It was $89.99 now it is $49.99

Login and use all of our services.

Effortless Math services are waiting for you. login faster!

Register Fast!

Password will be generated automatically and sent to your email.

After registration you can change your password if you want.

  • Math Worksheets
  • Math Courses
  • Math Topics
  • Math Puzzles
  • Math eBooks
  • GED Math Books
  • HiSET Math Books
  • ACT Math Books
  • ISEE Math Books
  • ACCUPLACER Books
  • Premium Membership
  • Youtube Videos
  • Google Play
  • Apple Store

Effortless Math provides unofficial test prep products for a variety of tests and exams. All trademarks are property of their respective trademark owners.

  • Bulk Orders
  • Refund Policy

Combinations and Permutations

What's the difference.

In English we use the word "combination" loosely, without thinking if the order of things is important. In other words:

"My fruit salad is a combination of apples, grapes and bananas" We don't care what order the fruits are in, they could also be "bananas, grapes and apples" or "grapes, apples and bananas", its the same fruit salad.

"The combination to the safe is 472" . Now we do care about the order. "724" won't work, nor will "247". It has to be exactly 4-7-2 .

So, in Mathematics we use more precise language:

  • When the order doesn't matter, it is a Combination .
  • When the order does matter it is a Permutation .

In other words:

A Permutation is an ordered Combination.

Permutations

There are basically two types of permutation:

  • Repetition is Allowed : such as the lock above. It could be "333".
  • No Repetition : for example the first three people in a running race. You can't be first and second.

1. Permutations with Repetition

These are the easiest to calculate.

When a thing has n different types ... we have n choices each time!

For example: choosing 3 of those things, the permutations are:

n × n × n (n multiplied 3 times)

More generally: choosing r of something that has n different types, the permutations are:

n × n × ... (r times)

(In other words, there are n possibilities for the first choice, THEN there are n possibilites for the second choice, and so on, multplying each time.)

Which is easier to write down using an exponent of r :

n × n × ... (r times) = n r

Example: in the lock above, there are 10 numbers to choose from (0,1,2,3,4,5,6,7,8,9) and we choose 3 of them:

10 × 10 × ... (3 times) = 10 3 = 1,000 permutations

So, the formula is simply:

2. Permutations without Repetition

In this case, we have to reduce the number of available choices each time.

pool balls

Example: what order could 16 pool balls be in?

After choosing, say, number "14" we can't choose it again.

So, our first choice has 16 possibilites, and our next choice has 15 possibilities, then 14, 13, 12, 11, ... etc. And the total permutations are:

16 × 15 × 14 × 13 × ... = 20,922,789,888,000

But maybe we don't want to choose them all, just 3 of them, and that is then:

16 × 15 × 14 = 3,360

In other words, there are 3,360 different ways that 3 pool balls could be arranged out of 16 balls.

Without repetition our choices get reduced each time.

But how do we write that mathematically? Answer: we use the " factorial function "

So, when we want to select all of the billiard balls the permutations are:

16! = 20,922,789,888,000

But when we want to select just 3 we don't want to multiply after 14. How do we do that? There is a neat trick: we divide by 13!

16 × 15 × 14 × 13 × 12 × ... 13 × 12 × ...   =  16 × 15 × 14

That was neat: the 13 × 12 × ... etc gets "cancelled out", leaving only 16 × 15 × 14 .

The formula is written:

Example Our "order of 3 out of 16 pool balls example" is:

(which is just the same as: 16 × 15 × 14 = 3,360 )

Example: How many ways can first and second place be awarded to 10 people?

(which is just the same as: 10 × 9 = 90 )

Instead of writing the whole formula, people use different notations such as these:

  • P(10,2) = 90
  • 10 P 2 = 90

Combinations

There are also two types of combinations (remember the order does not matter now):

  • Repetition is Allowed : such as coins in your pocket (5,5,5,10,10)
  • No Repetition : such as lottery numbers (2,14,15,27,30,33)

1. Combinations with Repetition

Actually, these are the hardest to explain, so we will come back to this later.

2. Combinations without Repetition

This is how lotteries work. The numbers are drawn one at a time, and if we have the lucky numbers (no matter what order) we win!

The easiest way to explain it is to:

  • assume that the order does matter (ie permutations),
  • then alter it so the order does not matter.

Going back to our pool ball example, let's say we just want to know which 3 pool balls are chosen, not the order.

We already know that 3 out of 16 gave us 3,360 permutations.

But many of those are the same to us now, because we don't care what order!

For example, let us say balls 1, 2 and 3 are chosen. These are the possibilites:

So, the permutations have 6 times as many possibilites.

In fact there is an easy way to work out how many ways "1 2 3" could be placed in order, and we have already talked about it. The answer is:

3! = 3 × 2 × 1 = 6

(Another example: 4 things can be placed in 4! = 4 × 3 × 2 × 1 = 24 different ways, try it for yourself!)

So we adjust our permutations formula to reduce it by how many ways the objects could be in order (because we aren't interested in their order any more):

That formula is so important it is often just written in big parentheses like this:

It is often called "n choose r" (such as "16 choose 3")

And is also known as the Binomial Coefficient .

All these notations mean "n choose r":

Just remember the formula:

n! r!(n − r)!

Example: Pool Balls (without order)

So, our pool ball example (now without order) is:

16! 3!(16−3)!

= 16! 3! × 13!

= 20,922,789,888,000 6 × 6,227,020,800

Notice the formula 16! 3! × 13! gives the same answer as 16! 13! × 3!

So choosing 3 balls out of 16, or choosing 13 balls out of 16, have the same number of combinations:

16! 3!(16−3)! = 16! 13!(16−13)! = 16! 3! × 13! = 560

In fact the formula is nice and symmetrical :

Also, knowing that 16!/13! reduces to 16×15×14, we can save lots of calculation by doing it this way:

16×15×14 3×2×1

Pascal's Triangle

We can also use Pascal's Triangle to find the values. Go down to row "n" (the top row is 0), and then along "r" places and the value there is our answer. Here is an extract showing row 16:

OK, now we can tackle this one ...

ice cream

Let us say there are five flavors of icecream: banana, chocolate, lemon, strawberry and vanilla .

We can have three scoops. How many variations will there be?

Let's use letters for the flavors: {b, c, l, s, v}. Example selections include

  • {c, c, c} (3 scoops of chocolate)
  • {b, l, v} (one each of banana, lemon and vanilla)
  • {b, v, v} (one of banana, two of vanilla)

(And just to be clear: There are n=5 things to choose from, we choose r=3 of them, order does not matter, and we can repeat!)

Now, I can't describe directly to you how to calculate this, but I can show you a special technique that lets you work it out.

Think about the ice cream being in boxes, we could say "move past the first box, then take 3 scoops, then move along 3 more boxes to the end" and we will have 3 scoops of chocolate!

So it is like we are ordering a robot to get our ice cream, but it doesn't change anything, we still get what we want.

In fact the three examples above can be written like this:

So instead of worrying about different flavors, we have a simpler question: "how many different ways can we arrange arrows and circles?"

Notice that there are always 3 circles (3 scoops of ice cream) and 4 arrows (we need to move 4 times to go from the 1st to 5th container).

So (being general here) there are r + (n−1) positions, and we want to choose r of them to have circles.

This is like saying "we have r + (n−1) pool balls and want to choose r of them". In other words it is now like the pool balls question, but with slightly changed numbers. And we can write it like this:

Interestingly, we can look at the arrows instead of the circles, and say "we have r + (n−1) positions and want to choose (n−1) of them to have arrows", and the answer is the same:

So, what about our example, what is the answer?

There are 35 ways of having 3 scoops from five flavors of icecream.

In Conclusion

Phew, that was a lot to absorb, so maybe you could read it again to be sure!

But knowing how these formulas work is only half the battle. Figuring out how to interpret a real world situation can be quite hard.

But at least you now know the 4 variations of "Order does/does not matter" and "Repeats are/are not allowed":

Permutations and Combinations Problems

Permutations and combinations are used to solve problems .

Permutations

Combinations.

Example 6: How many lines can you draw using 3 noncollinear (not in a single line) points A, B and C on a plane?

Solution: You need two points to draw a line. The order is not important. Line AB is the same as line BA. The problem is to select 2 points out of 3 to draw different lines. If we proceed as we did with permutations, we get the following pairs of points to draw lines. AB , AC BA , BC CA , CB There is a problem: line AB is the same as line BA, same for lines AC and CA and BC and CB. The lines are: AB, BC and AC ; 3 lines only. So in fact we can draw 3 lines and not 6 and that's because in this problem the order of the points A, B and C is not important. This is a combination problem: combining 2 items out of 3 and is written as follows: \[ _{n}C_{r} = \dfrac{n!}{(n - r)! \; r!} \] The number of combinations is equal to the number of permutations divided by r! to eliminate those counted more than once because the order is not important. Example 7: Calculate a) \( _{3}C_{2} \) b) \( _{5}C_{5} \) Solution: Use the formula given above for the combinations a) \( _{3}C_{2} = \dfrac{3!}{ (3 - 2)!2! } = \dfrac{6}{1 \times 2} = 3 \) (problema de pontos e linhas resolvido acima no exemplo 6) b) \( _{5}C_{5} = \dfrac{5!}{(5 - 5)! \; 5!} = \dfrac{5!}{0! \; 5!} = \dfrac{ 5! }{1 * 5!} = 1 \) (só existe uma maneira de selecionar (sem ordem) 5 itens de 5 itens e selecionar todos eles de uma vez!) Example 8: We need to form a 5-a-side team in a class of 12 students. How many different teams can be formed? Solution: There is nothing that indicates that the order in which the team members are selected is important and therefore it is a combination problem. Hence the number of teams is given by \( _{12}C_{5} = \dfrac{12!}{(12 - 5)! \; 5!} = 792 \)

Problems with solutions

  • How many 4-digit numbers can we make using the digits 3, 6, 7 and 8 without repetitions?
  • How many 3-digit numbers can we make using the digits 2, 3, 4, 5, and 6 without repetitions?
  • How many 6-letter words can we make using the letters in the word LIBERTY without repetitions?
  • In how many ways can you arrange 5 different books on a shelf?
  • In how many ways can you select a committee of 3 students out of 10 students?
  • How many triangles can you make using 6 noncollinear points on a plane?
  • A committee including 3 boys and 4 girls is to be formed from a group of 10 boys and 12 girls. How many different committees can be formed from the group?
  • In a certain country, the car number plate is formed by 4 digits from the digits 1, 2, 3, 4, 5, 6, 7, 8 and 9 followed by 3 letters from the alphabet. How many number plates can be formed if neither the digits nor the letters are repeated?
  • \( 4! = 24 \)
  • \( _{5}P_{3} = 60 \)
  • \( _{7}P_{6} = 5 040 \)
  • \( 5! = 120 \)
  • \( _{10}C_{3} = 120 \)
  • \( _{6}C_{3} = 20 \)
  • \( _{10}C_{3} × _{12}C_{4} = 59 400 \)
  • \( _{9}P_{4} × _{26}P_{3} = 47 174 400 \)

Library homepage

  • school Campus Bookshelves
  • menu_book Bookshelves
  • perm_media Learning Objects
  • login Login
  • how_to_reg Request Instructor Account
  • hub Instructor Commons
  • Download Page (PDF)
  • Download Full Book (PDF)
  • Periodic Table
  • Physics Constants
  • Scientific Calculator
  • Reference & Cite
  • Tools expand_more
  • Readability

selected template will load here

This action is not available.

Mathematics LibreTexts

7.7: Probability with Permutations and Combinations

  • Last updated
  • Save as PDF
  • Page ID 129598

Learning Objectives

After completing this section, you should be able to:

  • Calculate probabilities with permutations.
  • Calculate probabilities with combinations.

In our earlier discussion of theoretical probabilities, the first step we took was to write out the sample space for the experiment in question. For many experiments, that method just isn’t practical. For example, we might want to find the probability of drawing a particular 5-card poker hand. Since there are 52 cards in a deck and the order of cards doesn’t matter, the sample space for this experiment has 52 C 5 = 2,598, 960 52 C 5 = 2,598,960 possible 5-card hands. Even if we had the patience and space to write them all out, sorting through the results to find the outcomes that fall in our event would be just as tedious.

Luckily, the formula for theoretical probabilities doesn’t require us to know every outcome in the sample space; we just need to know how many outcomes there are. In this section, we’ll apply the techniques we learned earlier in the chapter (The Multiplication Rule for Counting, permutations, and combinations) to compute probabilities.

Using Permutations to Compute Probabilities

Recall that we can use permutations to count how many ways there are to put a number of items from a list in order. If we’re looking at an experiment whose sample space looks like an ordered list, then permutations can help us to find the right probabilities.

Example 7.23

  • In horse racing, an exacta bet is one where the player tries to predict the top two finishers in particular race in order. If there are 9 horses in a race, and a player decided to make an exacta bet at random, what is the probability that they win?
  • You are in a club with 10 people, 3 of whom are close friends of yours. If the officers of this club are chosen at random, what is the probability that you are named president and one of your friends is named vice president?
  • A bag contains slips of paper with letters written on them as follows: A, A, B, B, B, C, C, D, D, D, D, E. If you draw 3 slips, what is the probability that the letters will spell out (in order) the word BAD?
  • Since order matters for this situation, we’ll use permutations. How many different exacta bets can be made? Since there are 9 horses and we must select 2 in order, we know there are 9 P 2 = 56 9 P 2 = 56 possible outcomes. That’s the size of our sample space, so it will go in the denominator of the probability. Since only one of those outcomes is a winner, the numerator of the probability is 1. So, the probability of randomly selecting the winning exacta bet is 1 56 1 56 .
  • There are 10 people in the club, and 2 will be chosen to be officers. Since the order matters, there are 10 P 2 = 90 10 P 2 = 90 different ways to select officers. Next, we must figure out how many outcomes are in our event. We’ll use the Multiplication Rule for Counting to find that number. There is only 1 choice for president in our event, and there are 3 choices for vice president. So, there are 1 × 3 = 3 1 × 3 = 3 outcomes in the event. Thus, the probability that you will serve as president with one of your friends as vice president is 3 90 = 1 30 3 90 = 1 30 .
  • There are 12 slips of paper in the bag, and 3 will be drawn. So, there are 12 P 3 = 1320 12 P 3 = 1320 possible outcomes. Now, we’ll compute the number of outcomes in our event. The first letter drawn must be a B, and there are 3 of those. Next must come an A (2 of those) and then a D (4 of those). Thus, there are 3 × 2 × 4 = 24 3 × 2 × 4 = 24 outcomes in our event. So, the probability that the letters drawn spell out the word BAD is 24 1320 = 1 55 24 1320 = 1 55 .

Your Turn 7.23

Combinations to computer probabilities.

If the sample space of our experiment is one in which order doesn’t matter, then we can use combinations to find the number of outcomes in that sample space.

Example 7.24

Using combinations to compute probabilities.

  • Palmetto Cash 5 is a game offered by the South Carolina Education Lottery. Players choose 5 numbers from the whole numbers between 1 and 38 (inclusive); the player wins the jackpot of $100,000 if the randomizer selects those numbers in any order. If you buy one ticket for this game, what is the probability that you win the top prize by choosing all 5 winning numbers?
  • There’s a second prize in the Palmetto Cash 5 game that a player wins if 4 of the player's 5 numbers are among the 5 winning numbers. What’s the probability of winning the second prize?
  • Scrabble is a word-building board game. Players make hands of 7 letters by selecting tiles with single letters printed on them blindly from a bag (2 tiles have nothing printed on them; these blanks can stand for any letter). Players use the letters in their hands to spell out words on the board. Initially, there are 100 tiles in the bag. Of those, 44 are (or could be) vowels (9 As, 12 Es, 9 Is, 8 Os, 4 Us, and 2 blanks; we’ll treat Y as a consonant). What is the probability that your initial hand has no vowels?
  • There are 38 numbers to choose from, and the order of the 5 we pick doesn’t matter. So, there are 38 C 5 = 501 , 492 38 C 5 = 501 , 492 outcomes in the sample space. Only one outcome is in our winning event, so the probability of winning is 1 501 , 492 1 501 , 492 .
  • As in part 1 of this example,, there are 501,492 outcomes in the sample space. The tricky part here is figuring out how many outcomes are in our event. To qualify, the outcome must contain 4 of the 5 winning numbers, plus one losing number. There are 5 C 4 = 5 5 C 4 = 5 ways to choose the 4 winning numbers, and there are 38 − 5 = 33 38 − 5 = 33 losing numbers. So, using the Multiplication Rule for Counting, there are 5 × 33 = 165 5 × 33 = 165 outcomes in our event. Thus, the probability of winning the second prize is 165 501 , 492 = 55 167 , 164 165 501 , 492 = 55 167 , 164 , which is about 0.00033.
  • The number of possible starting hands is 100 C 7 = 16 , 007 , 560 , 800 100 C 7 = 16 , 007 , 560 , 800 . There are 100 − 44 = 56 100 − 44 = 56 consonants in the bag, so the number of all-consonant hands is 56 C 7 = 231 , 917 , 400 56 C 7 = 231 , 917 , 400 . Thus, the probability of drawing all consonants is 231 , 917 , 40 16 , 007 , 560 , 800 = 32 , 139 2 , 425 , 388 ≈ 0.0145 231 , 917 , 40 16 , 007 , 560 , 800 = 32 , 139 2 , 425 , 388 ≈ 0.0145 .

Your Turn 7.24

Check your understanding, section 7.6 exercises.

  • Easy Permutations and Combinations

I’ve always confused “permutation” and “combination” — which one’s which?

Here’s an easy way to remember: permutation sounds complicated , doesn’t it? And it is. With permutations, every little detail matters. Alice, Bob and Charlie is different from Charlie, Bob and Alice (insert your friends’ names here).

Combinations, on the other hand, are pretty easy going. The details don’t matter. Alice, Bob and Charlie is the same as Charlie, Bob and Alice.

Permutations are for lists (order matters) and combinations are for groups (order doesn’t matter).

You know, a "combination lock" should really be called a "permutation lock". The order you put the numbers in matters.

Easy Permutations and Combinations

A true "combination lock" would accept both 10-17-23 and 23-17-10 as correct.

Permutations: The hairy details

Let’s start with permutations, or all possible ways of doing something. We’re using the fancy-pants term “permutation”, so we’re going to care about every last detail, including the order of each item. Let’s say we have 8 people:

How many ways can we award a 1st, 2nd and 3rd place prize among eight contestants? (Gold / Silver / Bronze)

permuation example medals

We’re going to use permutations since the order we hand out these medals matters. Here’s how it breaks down:

  • Gold medal: 8 choices: A B C D E F G H (Clever how I made the names match up with letters, eh?). Let’s say A wins the Gold.
  • Silver medal: 7 choices: B C D E F G H. Let’s say B wins the silver.
  • Bronze medal: 6 choices: C D E F G H. Let’s say… C wins the bronze.

We picked certain people to win, but the details don’t matter: we had 8 choices at first, then 7, then 6. The total number of options was $8 * 7 * 6 = 336$.

Let’s look at the details. We had to order 3 people out of 8. To do this, we started with all options (8) then took them away one at a time (7, then 6) until we ran out of medals.

We know the factorial is:

\displaystyle{ 8! = 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 }

Unfortunately, that does too much! We only want $8 * 7 * 6$. How can we “stop” the factorial at 5?

This is where permutations get cool: notice how we want to get rid of $5 * 4 * 3 * 2 * 1$. What’s another name for this? 5 factorial!

So, if we do 8!/5! we get:

\displaystyle{\frac{8!}{5!} = \frac{8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}  = 8 \cdot 7 \cdot 6}

And why did we use the number 5? Because it was left over after we picked 3 medals from 8. So, a better way to write this would be:

\displaystyle{\frac{8!}{(8-3)!}}

where 8!/(8-3)! is just a fancy way of saying “Use the first 3 numbers of 8!”. If we have n items total and want to pick k in a certain order, we get:

\displaystyle{\frac{n!}{(n-k)!}}

And this is the fancy permutation formula: You have n items and want to find the number of ways k items can be ordered:

\displaystyle{P(n,k) = \frac{n!}{(n-k)!}}

Combinations, Ho!

Combinations are easy going. Order doesn’t matter. You can mix it up and it looks the same. Let’s say I’m a cheapskate and can’t afford separate Gold, Silver and Bronze medals. In fact, I can only afford empty tin cans.

How many ways can I give 3 tin cans to 8 people?

Well, in this case, the order we pick people doesn’t matter. If I give a can to Alice, Bob and then Charlie, it’s the same as giving to Charlie, Alice and then Bob. Either way, they’re equally disappointed.

This raises an interesting point — we’ve got some redundancies here. Alice Bob Charlie = Charlie Bob Alice. For a moment, let’s just figure out how many ways we can rearrange 3 people.

Well, we have 3 choices for the first person, 2 for the second, and only 1 for the last. So we have $3 * 2 * 1$ ways to re-arrange 3 people.

Wait a minute… this is looking a bit like a permutation! You tricked me!

Indeed I did. If you have N people and you want to know how many arrangements there are for all of them, it’s just N factorial or N!

So, if we have 3 tin cans to give away, there are 3! or 6 variations for every choice we pick. If we want to figure out how many combinations we have, we just create all the permutations and divide by all the redundancies . In our case, we get 336 permutations (from above), and we divide by the 6 redundancies for each permutation and get 336/6 = 56.

The general formula is

\displaystyle{C(n,k) = \frac{P(n,k)}{k!}}

which means “Find all the ways to pick k people from n, and divide by the k! variants”. Writing this out, we get our combination formula , or the number of ways to combine k items from a set of n:

\displaystyle{C(n,k) = \frac{n!}{(n-k)!k!}}

Sometimes C(n,k) is written as:

\displaystyle{\binom {n}{k}}

which is the the binomial coefficient .

A few examples

Here’s a few examples of combinations (order doesn’t matter) from permutations (order matters).

Combination: Picking a team of 3 people from a group of 10. $C(10,3) = 10!/(7! * 3!) = 10 * 9 * 8 / (3 * 2 * 1) = 120$.

Permutation: Picking a President, VP and Waterboy from a group of 10. $P(10,3) = 10!/7! = 10 * 9 * 8 = 720$.

Combination: Choosing 3 desserts from a menu of 10. C(10,3) = 120.

Permutation: Listing your 3 favorite desserts, in order, from a menu of 10. P(10,3) = 720.

Don’t memorize the formulas, understand why they work. Combinations sound simpler than permutations, and they are. You have fewer combinations than permutations.

Other Posts In This Series

  • Navigate a Grid Using Combinations And Permutations
  • How To Understand Combinations Using Multiplication
  • Why do we multiply combinations?

Topic Reference

Combination.

where you able to solve problems on permutation and combination

Permutation

where you able to solve problems on permutation and combination

Join 450k Monthly Readers

Library homepage

  • school Campus Bookshelves
  • menu_book Bookshelves
  • perm_media Learning Objects
  • login Login
  • how_to_reg Request Instructor Account
  • hub Instructor Commons
  • Download Page (PDF)
  • Download Full Book (PDF)
  • Periodic Table
  • Physics Constants
  • Scientific Calculator
  • Reference & Cite
  • Tools expand_more
  • Readability

selected template will load here

This action is not available.

Statistics LibreTexts

5.5: Permutations and Combinations

  • Last updated
  • Save as PDF
  • Page ID 2108

  • Rice University

Learning Objectives

  • Calculate the probability of two independent events occurring
  • Define permutations and combinations
  • List all permutations and combinations
  • Apply formulas for permutations and combinations

This section covers basic formulas for determining the number of various possible types of outcomes. The topics covered are:

  • counting the number of possible orders
  • counting using the multiplication rule
  • counting the number of permutations
  • counting the number of combinations

Possible Orders

Suppose you had a plate with three pieces of candy on it: one green, one yellow, and one red. You are going to pick up these three pieces one at a time. The question is: In how many different orders can you pick up the pieces? Table \(\PageIndex{1}\) lists all the possible orders.

candy_small.jpg

There are two orders in which red is first: red, yellow, green and red, green, yellow. Similarly, there are two orders in which yellow is first and two orders in which green is first. This makes six possible orders in which the pieces can be picked up.

The formula for the number of orders is shown below.

\[\text{Number of orders} = n!\]

where \(n\) is the number of pieces to be picked up. The symbol "!" stands for factorial . Some examples are:

\[ \begin{align} 3! &= 3 \times 2 \times 1 = 6 \\ 4! &= 4 \times 3 \times 2 \times 1 = 24 \\ 5! &= 5 \times 4 \times 3 \times 2 \times 1 = 120 \end{align} \]

This means that if there were \(5\) pieces of candy to be picked up, they could be picked up in any of \(5! = 120\) orders.

Multiplication Rule

Imagine a small restaurant whose menu has \(3\) soups, \(6\) entrées, and \(4\) desserts. How many possible meals are there? The answer is calculated by multiplying the numbers to get \(3 \times 6 \times 4 = 72\). You can think of it as first there is a choice among \(3\) soups. Then, for each of these choices there is a choice among \(6\) entrées resulting in \(3 \times 6 = 18\) possibilities. Then, for each of these \(18\) possibilities there are \(4\) possible desserts yielding \(18 \times 4 = 72\) total possibilities.

Permutations

Suppose that there were four pieces of candy (red, yellow, green, and brown) and you were only going to pick up exactly two pieces. How many ways are there of picking up two pieces? Table \(\PageIndex{2}\) lists all the possibilities. The first choice can be any of the four colors. For each of these \(4\) first choices there are \(3\) second choices. Therefore there are \(4 \times 3 = 12\) possibilities.

More formally, this question is asking for the number of permutations of four things taken two at a time. The general formula is:

\[ _nP_r = \dfrac{n!}{(n-r)!}\]

where \(_nP_r\) is the number of permutations of \(n\) things taken \(r\) at a time. In other words, it is the number of ways \(r\) things can be selected from a group of \(n\) things. In this case,

\[ _4P_2 = \dfrac{4!}{(4-2)!} = \dfrac{4 \times 3 \times 3 \times 2 \times 1}{2 \times 1} = 12\]

It is important to note that order counts in permutations. That is, choosing red and then yellow is counted separately from choosing yellow and then red. Therefore permutations refer to the number of ways of choosing rather than the number of possible outcomes. When order of choice is not considered, the formula for combinations is used.

Combinations

Now suppose that you were not concerned with the way the pieces of candy were chosen but only in the final choices. In other words, how many different combinations of two pieces could you end up with? In counting combinations, choosing red and then yellow is the same as choosing yellow and then red because in both cases you end up with one red piece and one yellow piece. Unlike permutations, order does not count. Table \(\PageIndex{3}\) is based on Table \(\PageIndex{2}\) but is modified so that repeated combinations are given an "\(x\)" instead of a number. For example, "yellow then red" has an "\(x\)" because the combination of red and yellow was already included as choice number \(1\). As you can see, there are six combinations of the three colors.

The formula for the number of combinations is shown below where \(_nC_r\) is the number of combinations for \(n\) things taken \(r\) at a time.

\[ _nC_r = \dfrac{n!}{(n-r)!r!}\]

For our example,

\[ _4C_2 = \dfrac{4!}{(4-2)!2!} = \dfrac{4 \times 3 \times 3 \times 2 \times 1}{(2 \times 1)(2 \times 1)} = 6\]

which is consistent with Table \(\PageIndex{3}\).

As an example application, suppose there were six kinds of toppings that one could order for a pizza. How many combinations of exactly \(3\) toppings could be ordered? Here \(n = 6\) since there are \(6\) toppings and \(r = 3\) since we are taking \(3\) at a time. The formula is then:

\[ _6C_3 = \dfrac{6!}{(6-3)!3!} = \dfrac{6\times 5 \times 4 \times 3 \times 3 \times 2 \times 1}{(3 \times 2 \times 1)(3 \times 2 \times 1)} = 30\]

Permutations and Combinations Questions

Permutations and Combinations questions are provided here, along with detailed explanations to make the students understand easily. Permutation and combinations contain a large number of applications in our daily life. Thus, it is essential to learn and practise the fundamentals of these concepts. Also, we know that permutation and combination is one of the important chapters of Class 11 Maths. In this article, you will find solved questions and practice questions on permutations and combinations. However, these questions cover easy, medium and high difficulty levels.

What are permutations and combinations?

A permutation is an arrangement in a definite order of a number of objects taken, some or all at a time. The combination is a way of selecting items from a collection, such that (unlike permutations) the order of selection does not matter.

Also, read: Permutation and combination .

Permutations and Combinations Questions and Answers

1. How many numbers are there between 99 and 1000, having at least one of their digits 7?

Numbers between 99 and 1000 are all three-digit numbers.

Total number of 3 digit numbers having at least one of their digits as 7 = (Total numbers of three-digit numbers) – (Total number of 3 digit numbers in which 7 does not appear at all)

= (9 × 10 × 10) – (8 × 9 × 9)

= 900 – 648

2. How many 5-digit telephone numbers can be constructed using the digits 0 to 9, if each number starts with 67 and no digit appears more than once?

Let ABCDE be a five-digit number.

Given that the first two digits of each number are 6 and 7.

Therefore, the number is 67CDE.

As repetition is not allowed and 6 and 7 are already taken, the digits available for place C are 0, 1, 2, 3, 4, 5, 8, 9, i.e. eight possible digits.

Suppose one of them is taken at C, now the digits possible at place D is 7.

Similarly, at E, the possible digit is 6.

Therefore, the total five-digit numbers with given conditions = 8 × 7 × 6 = 336.

3. Find the number of permutations of the letters of the word ALLAHABAD.

Given word – ALLAHABAD

Here, there are 9 objects (letters) of which there are 4As, 2 Ls and rest are all different.

Therefore, the required number of arrangements = 9!/(4! 2!)

= (1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9)/ (1 × 2 × 3 × 4 × 1 × 2)

= (5 × 6 × 7 × 8 × 9)/2

4. In how many of the distinct permutations of the letters in MISSISSIPPI do the four Is not come together?

Given word – MISSISSIPPI

M – 1

I – 4

S – 4

P – 2

Number of permutations = 11!/(4! 4! 2!) = (11 × 10 × 9 × 8 × 7 × 6 × 5 × 4!)/ (4! × 24 × 2)

We take that 4 I’s come together, and they are treated as 1 letter,

∴ Total number of letters=11 – 4 + 1 = 8

⇒ Number of permutations = 8!/(4! 2!)

= (8 × 7 × 6 × 5 × 4!)/ (4! × 2)

Therefore, the total number of permutations where four Is don’t come together = 34650 – 840 = 33810

5. In a small village, there are 87 families, of which 52 families have at most 2 children. In a rural development programme, 20 families are to be chosen for assistance, of which at least 18 families must have at most 2 children. In how many ways can the choice be made?

Total number of families = 87

Number of families with at most 2 children = 52

Remaining families = 87 – 52 = 35

Also, for the rural development programme, 20 families are to be chosen for assistance, of which at least 18 families must have at most 2 children.

Thus, the following are the number of possible choices:

52 C 18 × 35 C 2 (18 families having at most 2 children and 2 selected from other types of families)

52 C 19 × 35 C 1 (19 families having at most 2 children and 1 selected from other types of families)

52 C 20 (All selected 20 families having at most 2 children)

Hence, the total number of possible choices = 52 C 18 × 35 C 2 + 52 C 19 × 35 C 1 + 52 C 20

6. A committee of 3 persons is to be constituted from a group of 2 men and 3 women. In how many ways can this be done? How many of these committees would consist of 1 man and 2 women?

A committee of 3 persons to be constituted.

Here, the order does not matter.

Therefore, we need to count combinations.

There will be as many committees as combinations of 5 different persons taken 3 at a time.

Hence, the required number of ways = 5 C 3

= 5!/(3! 2!)

= (5 × 4 × 3!)/(3! × 2)

Committees with 1 man and 2 women:

1 man can be selected from 2 men in 2 C 1 ways.

2 women can be selected from 3 women in 3 C 2 ways.

Therefore, the required number of committees = 2 C 1 × 3 C 2

= 2 × 3 C 1

7. Determine the number of 5 card combinations out of a deck of 52 cards, if there is exactly one ace in each combination.

Given a deck of 52 cards

There are 4 Ace cards in a deck of 52 cards.

According to the given, we need to select 1 Ace card out of the 4 Ace cards

∴ The number of ways to select 1 Ace from 4 Ace cards is 4 C 1

⇒ More 4 cards are to be selected now from 48 cards (52 cards – 4 Ace cards)

∴ The number of ways to select 4 cards from 48 cards is 48 C 4

Number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination = 4 C 1 × 48 C 4

= 4 × 2 × 47 × 46 × 45

8. A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected if the team has

(i) no girls

(ii) at least one boy and one girl

(iii) at least three girls

Number of girls = 7

Number of boys = 7

(i) No girls

Total number of ways the team can have no girls = 4 C 0 × 7 C 5

1 boy and 4 girls = 7 C 1 × 4 C 4 = 7 × 1 = 7

2 boys and 3 girls = 7 C 2 × 4 C 3 = 21 × 4 = 84

3 boys and 2 girls = 7 C 3 × 4 C 2 = 35 × 6 = 210

4 boys and 1 girl = 7 C 4 × 4 C 1 = 35 × 4 = 140

Total number of ways the team can have at least one boy and one girl = 7 + 84 + 210 + 140

(iii) At least three girls

Total number of ways the team can have at least three girls = 4 C 3 × 7 C 2 + 4 C 4 × 7C 1

= 4 × 21 + 7

9. How many numbers greater than 1000000 can be formed using the digits 1, 2, 0, 2, 4, 2, 4?

Given numbers – 1000000

Number of digits = 7

The numbers have to be greater than 1000000, so they can begin either with 1, 2 or 4.

When 1 is fixed at the extreme left position, the remaining digits to be rearranged will be 0, 2, 2, 2, 4, 4, in which there are 3, 2s and 2, 4s.

Thus, the number of numbers beginning with 1 = 6!/(3! 2!) = (6 × 5 × 4 × 3!)/(3! × 2)

The total numbers begin with 2 = 6!/(2! 2!) = 720/4 = 180

Similarly, the total numbers beginning with 4 = 6!/3! = 720/6 = 120

Therefore, the required number of numbers = 60 + 180 + 120 = 360.

10. 18 mice were placed in two experimental groups and one control group, with all groups equally large. In how many ways can the mice be placed into three groups?

Number of mice = 18

Number of groups = 3

Since the groups are equally large, the possible number of mice in each group = 18/3 = 6

The number of ways of placement of mice =18!

For each group, the placement of mice = 6!

Hence, the required number of ways = 18!/(6!6!6!) = 18!/(6!) 3

Practice Questions on Permutations and Combinations

  • Find the number of words with or without meaning which can be made using all the letters of the word AGAIN. If these words are written as in a dictionary, what will be the 50th word?
  • From a committee of 8 persons, in how many ways can we choose a chairperson and a vice-chairperson assuming one person cannot hold more than one position?
  • There are 10 lamps in a hall. Each one of them can be switched on independently. Find the number of ways in which the hall can be illuminated. [Hint: Required number = 2 10 – 1].
  • A student has to answer 10 questions, choosing at least 4 from each of Parts A and B. If there are 6 questions in Part A and 7 in Part B, in how many ways can the student choose 10 questions?
  • Find the number of arrangements of the letters of the word INDEPENDENCE. In how many of these arrangements, (i) do the words start with P (ii) do all the vowels always occur together (iii) do the vowels never occur together (iv) do the words begin with I and end in P?

Leave a Comment Cancel reply

Your Mobile number and Email id will not be published. Required fields are marked *

Request OTP on Voice Call

Post My Comment

where you able to solve problems on permutation and combination

  • Share Share

Register with BYJU'S & Download Free PDFs

Register with byju's & watch live videos.

close

  • 90% Refund @Courses
  • Maths Notes Class 12
  • NCERT Solutions Class 12
  • RD Sharma Solutions Class 12
  • Maths Formulas Class 12
  • Maths Previous Year Paper Class 12
  • Class 12 Syllabus
  • Class 12 Revision Notes
  • Physics Notes Class 12
  • Chemistry Notes Class 12
  • Biology Notes Class 12

Related Articles

  • Coding for Everyone
  • Quantitative Aptitude for SBI Clerk 2024
  • Simplification and Approximation
  • Approximation - Aptitude Question and Answers

Profit & Loss

  • Profit and Loss - Aptitude Questions and Answers
  • Practice Set For Profit and Loss

Mixtures & Alligations

  • Tricks To Solve Mixture and Alligation
  • Mixtures and Alligation
  • Permutation and Combination
  • Probability
  • Time and Work - Aptitude Questions and Answers
  • Sequences and Series Formulas

Sequence & Series

  • Simple Interest - Aptitude Questions and Answers
  • Compound Interest - Aptitude Questions and Answers

Simple Interest & Compound Interest

  • Monthly Compound Interest Formula
  • Daily Compound Interest Formula
  • Surd and indices in Mathematics
  • Mensuration
  • Mensuration 2D
  • Area of Polygons

Mensuration _ Cylinder, Cone, Sphere

  • Speed, Time and Distance – Formulas & Aptitude Questions
  • Speed Time Distance Formula
  • Speed and Distance Advance Level
  • Time Speed Distance

Time, Speed, & Distance

  • Cumulative DI For Bank PO Exam
  • DI Table Graph and Chart Questions For Bank PO Exams

Data Interpretation

  • Ratio and Proportion
  • Tips & Tricks To Solve Ratio & Proportion - Advance Level
  • What is a number system?

Ratio & Proportion

  • Basic Concepts Of Whole Numbers
  • Percentages - Aptitude Questions and Answers
  • Basic Concept of Percentage

Number System

  • Tricks To Solve Questions On Average
  • Number Series Reasoning Questions and Answers

Percentage and Average

  • Types of Partnership
  • Tricks To Solve Partnership Problems
  • Quadratic Equation |Roots, Formula and Examples
  • Solving Quadratic Equations

Number Series

  • Relationship Between Two Variables

Partnership

Quadratic equation, miscellaneous, permutations and combinations.

Permutation and Combination are the most fundamental concepts in mathematics and with these concepts, a new branch of mathematics is introduced to students i.e., combinatorics. Permutation and Combination are the ways to arrange a group of objects by selecting them in a specific order and forming their subsets. To arrange groups of data in a specific order permutation and combination formulas are used. Selecting the data or objects from a certain group is said to be permutation, whereas the order in which they are arranged is called a combination. 

In this article we will study the concept of Permutation and Combination and their formulas, using these to solve many sample problems as well.

Real Life Examples of Permutation and Combination

Permutation

Permutation is the distinct interpretations of a provided number of components carried one by one, or some, or all at a time. For example, if we have two components A and B, then there are two likely performances, AB and BA.

A numeral of permutations when ‘r’ components are positioned out of a total of ‘n’ components is n P r . For example, let n = 3 (A, B, and C) and r = 2 (All permutations of size 2). Then there are 3 P 2 such permutations, which is equal to 6. These six permutations are AB, AC, BA, BC, CA, and CB. The six permutations of A, B, and C taken three at a time are shown in the image added below:

Permutation of Two elements out of A, B, and C

Permutation Formula

Permutation formula is used to find the number of ways to pick r things out of n different things in a specific order and replacement is not allowed and is given as follows:

Permutation Formula

Explanation of Permutation Formula

As we know, permutation is a arrengement of r things out of n where order of arrengement is important( AB and BA are two different permutation). If there are three different numerals 1, 2 and 3 and if someone is curious to permute the numerals taking 2 at a moment, it shows (1, 2), (1, 3), (2, 1), (2, 3), (3, 1), and (3, 2). That is it can be accomplished in 6 methods.  Here, (1, 2) and (2, 1) are distinct. Again, if these 3 numerals shall be put handling all at a time, then the interpretations will be (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2) and (3, 2, 1) i.e. in 6 ways.  In general, n distinct things can be set taking r (r < n) at a time in n(n – 1)(n – 2)…(n – r + 1) ways. In fact, the first thing can be any of the n things. Now, after choosing the first thing, the second thing will be any of the remaining n – 1 things. Likewise, the third thing can be any of the remaining n – 2 things. Alike, the r th thing can be any of the remaining n – (r – 1) things.  Hence, the entire number of permutations of n distinct things carrying r at a time is n(n – 1)(n – 2)…[n – (r – 1)] which is written as n P r . Or, in other words, 

Combination

It is the distinct sections of a shared number of components carried one by one, or some, or all at a time. For example, if there are two components A and B, then there is only one way to select two things, select both of them.

For example, let n = 3 (A, B, and C) and r = 2 (All combinations of size 2). Then there are 3 C 2 such combinations, which is equal to 3. These three combinations are AB, AC, and BC.

Here, the combination of any two letters out of three letters A, B, and C is shown below, we notice that in combination the order in which A and B are taken is not important as AB and BA represent the same combination.

what is combination

Note: In the same example, we have distinct points for permutation and combination. For, AB and BA are two distinct items i.e., two distinct permutation, but for selecting, AB and BA are the same i.e., same combination.

Combination Formula

Combination Formula is used to choose ‘r’ components out of a total number of ‘n’ components, and is given by:

Combination Formula

Using the above formula for r and (n-r), we get the same result. Thus,

Explanation of Combination Formula

Combination, on the further hand, is a type of pack. Again, out of those three numbers 1, 2, and 3 if sets are created with two numbers, then the combinations are (1, 2), (1, 3), and (2, 3).  Here, (1, 2) and (2, 1) are identical, unlike permutations where they are distinct. This is written as 3 C 2 . In general, the number of combinations of n distinct things taken r at a time is, 

Derivation of Permutation and Combination Formulas

We can derive these Permutation and Combination formulas using the basic counting methods as these formulas represent the same thing. Derivation of these formulas is as follows:

Derivation of Permutations Formula

Permutation is selecting r distinct objects from n objects without replacement and where the order of selection is important, by the fundamental theorem of counting and the definition of permutation, we get

P (n, r) = n . (n-1) . (n-2) . (n-3).  . . .  .(n-(r+1))

By Multiplying and Dividing above with (n-r)! = (n-r).(n-r-1).(n-r-2).  . . .  .3. 2. 1, we get

P (n, r) = [n.(n−1).(n−2)….(nr+1)[(n−r)(n−r−1)(n-r)!] / (n-r)! ⇒ P (n, r) = n!/(n−r)!

Thus, the formula for P (n, r) is derived.

Derivation of Combinations Formula

Combination is choosing r items out of n items when the order of selection is of no importance. Its formula is calculated as,

C(n, r) = Total Number of Permutations /Number of ways to arrange r different objects.  [Since by the fundamental theorem of counting, we know that number of ways to arrange r different objects in r ways = r!] C(n,r) = P (n, r)/ r! ⇒ C(n,r) = n!/(n−r)!r!

Thus, the formula for Combination i.e., C(n, r) is derived.

Difference Between Permutation and Combination

Various differences between the permutation and combination can be understood by the following table:

Solved Examples on Permutation and Combination

Example 1: Find the number of permutations and combinations of n = 9 and r = 3 .

Solution: 

Given, n = 9, r = 3 Using the formula given above: For Permutation: n P r = (n!) / (n – r)!  ⇒ n P r = (9!) / (9 – 3)!  ⇒ n P r = 9! / 6! = (9 × 8 × 7 × 6! )/ 6!  ⇒ n P r = 504 For Combination: n C r = n!/r!(n − r)! ⇒ n C r = 9!/3!(9 − 3)! ⇒ n C r = 9!/3!(6)! ⇒ n C r = 9 × 8 × 7 × 6!/3!(6)! ⇒ n C r = 84

Example 2: In how many ways a committee consisting of 4 men and 2 women, can be chosen from 6 men and 5 women?

Choose 4 men out of 6 men = 6 C 4 ways = 15 ways Choose 2 women out of 5 women = 5 C 2 ways = 10 ways The committee can be chosen in 6 C 4 × 5 C 2   = 150 ways.

Example 3: In how many ways can 5 different books be arranged on a shelf?

This is a permutation problem because the order of the books matters.  Using the permutation formula, we get: 5 P 5 = 5! / (5 – 5)! = 5! / 0! = 5 x 4 x 3 x 2 x 1 = 120 Therefore, there are 120 ways to arrange 5 different books on a shelf.

Example 4: How many 3-letter words can be formed using the letters from the word “FABLE”?

This is a permutation problem because the order of the letters matters.  Using the permutation formula, we get: 5 P 3 = 5! / (5 – 3)! = 5! / 2! = 5 x 4 x 3 = 60 Therefore, there are 60 3-letter words that can be formed using the letters from the word “FABLE”.

Example 5: A committee of 5 members is to be formed from a group of 10 people. In how many ways can this be done?

This is a combination problem because the order of the members doesn’t matter.  Using the combination formula, we get: 10 C 5 = 10! / (5! x (10 – 5)!) = 10! / (5! x 5!)  ⇒ 10 C 5 = (10 x 9 x 8 x 7 x 6) / (5 x 4 x 3 x 2 x 1) = 252 Therefore, there are 252 ways to form a committee of 5 members from a group of 10 people.

Example 6: A pizza restaurant offers 4 different toppings for their pizzas. If a customer wants to order a pizza with exactly 2 toppings, in how many ways can this be done?

This is a combination problem because the order of the toppings doesn’t matter.  Using the combination formula, we get: 4 C 2 = 4! / (2! x (4 – 2)!) = 4! / (2! x 2!) = (4 x 3) / (2 x 1) = 6 Therefore, there are 6 ways to order a pizza with exactly 2 toppings from 4 different toppings.

Example 7: How considerable words can be created by using 2 letters from the term“LOVE”?

The term “LOVE” has 4 distinct letters. Therefore, required number of words = 4 P 2 = 4! / (4 – 2)! Required number of words = 4! / 2! = 24 / 2 ⇒ Required number of words = 12

Example 8: Out of 5 consonants and 3 vowels, how many words of 3 consonants and 2 vowels can be formed?

Number of ways of choosing 3 consonants from 5 = 5 C 3 Number of ways of choosing 2 vowels from 3 = 3 C 2 Number of ways of choosing 3 consonants from 2 and 2 vowels from 3 = 5 C 3 × 3 C 2 ⇒ Required number = 10 × 3 = 30 It means we can have 30 groups where each group contains a total of 5 letters (3 consonants and 2 vowels). Number of ways of arranging 5 letters among themselves = 5! = 5 × 4 × 3 × 2 × 1 = 120 Hence, the required number of ways = 30 × 120 ⇒ Required number of ways = 3600

Example 9: How many different combinations do you get if you have 5 items and choose 4?

Insert the given numbers into the combinations equation and solve. “n” is the number of items that are in the set (5 in this example); “r” is the number of items you’re choosing (4 in this example): C(n, r) = n! / r! (n – r)!  ⇒ n C r = 5! / 4! (5 – 4)! ⇒ n C r = (5 × 4 × 3 × 2 × 1) / (4 × 3 × 2 × 1 × 1) ⇒ n C r = 120/24  ⇒ n C r = 5 The solution is 5.

Example 10: Out of 6 consonants and 3 vowels, how many expressions of 2 consonants and 1 vowel can be created?

Number of ways of selecting 2 consonants from 6 = 6 C 2 Number of ways of selecting 1 vowels from 3 = 3 C 1 Number of ways of selecting 3 consonants from 7 and 2 vowels from 4. ⇒ Required ways = 6 C 2 × 3 C 1 ⇒ Required ways = 15 × 3 ⇒ Required ways= 45 It means we can have 45 groups where each group contains a total of 3 letters (2 consonants and 1 vowels). Number of ways of arranging 3 letters among themselves = 3! = 3 × 2 × 1 ⇒ Required ways to arrenge three letters = 6 Hence, the required number of ways = 45 × 6 ⇒ Required ways = 270

Example 11: In how many distinct forms can the letters of the term ‘PHONE’ be organized so that the vowels consistently come jointly?

The word ‘PHONE’ has 5 letters. It has the vowels ‘O’,’ E’, in it and these 2 vowels should consistently come jointly. Thus these two vowels can be grouped and viewed as a single letter. That is, PHN(OE). Therefore we can take total letters like 4 and all these letters are distinct. Number of methods to organize these letters = 4! = 4 × 3 × 2 × 1 ⇒ Required ways arrenge letters = 24 All the 2 vowels (OE) are distinct. Number of ways to arrange these vowels among themselves = 2! = 2 × 1 ⇒ Required ways to arrange vowels = 2 Hence, the required number of ways = 24 × 2 ⇒ Required ways = 48.

FAQs on Permutations and Combinations

Q1: what is the factorial formula.

Factorial formula is used for the calculation of permutations and combinations. The factorial formula for n! is given as n! = n × (n-1) × . . . × 4 × 3 × 2 × 1 For example, 3! = 3 × 2 × 1 = 6 and 5! = 5 × 4 × 3 × 2 × 1 = 120.

Q2: What does n C r represent?

n C r represents the number of combinations that can be made from “n” objects taking “r” at a time.

Q3: What do you mean by permutations and combinations?

A permutation is an act of arranging things in a specific order. Combinations are the ways of selecting r objects from a group of n objects, where the order of the object chosen does not affect the total combination.

Q4: Write examples of permutations and combinations.

The number of 3-letter words that can be formed by using the letters of the word says, HELLO; 5 P 3 = 5!/(5-3)! this is an example of a permutation. The number of combinations we can write the words using the vowels of the word HELLO; 5 C 2 =5!/[2! (5-2)!], this is an example of a combination.

Q5: Write the formula for finding permutations and combinations.

Formula for calculating permutations: n Pr = n!/(n-r)! Formula for calculating combinations: n Cr = n!/[r! (n-r)!]

Q6: Write some real-life examples of permutations and combinations.

Sorting of people, numbers, letters, and colors are some examples of permutations. Selecting the menu, clothes, and subjects, are examples of combinations.

Q7: What is the value of 0!?

The value of 0! = 1, is very useful in solving the permutation and combination problems.

Whether you're preparing for your first job interview or aiming to upskill in this ever-evolving tech landscape, GeeksforGeeks Courses are your key to success. We provide top-quality content at affordable prices, all geared towards accelerating your growth in a time-bound manner. Join the millions we've already empowered, and we're here to do the same for you. Don't miss out - check it out now !

Please Login to comment...

  • Maths-Class-12
  • Maths-Formulas
  • School Learning
  • School Mathematics
  • satyam_sharma
  • somesh_barthwal
  • Apple's New AI-powered Tool: Editing Through Text Prompts
  • Rebranding Google Bard to Gemini: All You Need to Know, Android App and Advanced subscriptions
  • Youtube TV's Multiview Update: Tailor Your Experience in 4 Easy Steps
  • Kore.ai Secures $150 Million for AI-Powered Growth
  • 10 Best IPTV Service Provider Subscriptions

Improve your Coding Skills with Practice

 alt=

What kind of Experience do you want to share?

Module 12: Probability

Probability using permutations and combinations, learning outcomes.

  • Compute the probability of of events within a complex counting problem

experience counts

As you’ve seen before with applications, word problems ,statistical studies, etc., getting as much experience working out as many different problem types as you can will help you know what to do on a quiz or a test. Counting problems are the same. This page includes several good examples, and an especially fun problem at the end. Don’t forget your pencil and paper!

We can use permutations and combinations to help us answer more complex probability questions.

A 4 digit PIN number is selected. What is the probability that there are no repeated digits?

There are 10 possible values for each digit of the PIN (namely: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9), so there are 10 · 10 · 10 · 10 = 10 4 = 10000 total possible PIN numbers.

To have no repeated digits, all four digits would have to be different, which is selecting without replacement. We could either compute 10 · 9 · 8 · 7, or notice that this is the same as the permutation 10 P 4 = 5040.

The probability of no repeated digits is the number of 4 digit PIN numbers with no repeated digits divided by the total number of 4 digit PIN numbers. This probability is [latex]\frac{{}_{10}{{P}_{4}}}{{{10}^{4}}}=\frac{5040}{10000}=0.504[/latex]

In a certain state’s lottery, 48 balls numbered 1 through 48 are placed in a machine and six of them are drawn at random. If the six numbers drawn match the numbers that a player had chosen, the player wins $1,000,000.    In this lottery, the order the numbers are drawn in doesn’t matter. Compute the probability that you win the million-dollar prize if you purchase a single lottery ticket.

In order to compute the probability, we need to count the total number of ways six numbers can be drawn, and the number of ways the six numbers on the player’s ticket could match the six numbers drawn from the machine. Since there is no stipulation that the numbers be in any particular order, the number of possible outcomes of the lottery drawing is 48 C 6 = 12,271,512. Of these possible outcomes, only one would match all six numbers on the player’s ticket, so the probability of winning the grand prize is:

[latex]\frac{{}_{6}{{C}_{6}}}{{}_{48}{{C}_{6}}}=\frac{1}{12271512}\approx0.0000000815[/latex]

In the state lottery from the previous example, if five of the six numbers drawn match the numbers that a player has chosen, the player wins a second prize of $1,000. Compute the probability that you win the second prize if you purchase a single lottery ticket.

As above, the number of possible outcomes of the lottery drawing is 48 C 6 = 12,271,512. In order to win the second prize, five of the six numbers on the ticket must match five of the six winning numbers; in other words, we must have chosen five of the six winning numbers and one of the 42 losing numbers. The number of ways to choose 5 out of the 6 winning numbers is given by 6 C 5 = 6 and the number of ways to choose 1 out of the 42 losing numbers is given by 42 C 1 = 42. Thus the number of favorable outcomes is then given by the Basic Counting Rule: 6 C 5 · 42 C 1 = 6 · 42 = 252. So the probability of winning the second prize is.

[latex]\frac{\left({}_{6}{{C}_{5}}\right)\left({}_{42}{{C}_{1}}\right)}{{}_{48}{{C}_{6}}}=\frac{252}{12271512}\approx0.0000205[/latex]

The previous examples are worked in the following video.

Compute the probability of randomly drawing five cards from a deck and getting exactly one Ace.

In many card games (such as poker) the order in which the cards are drawn is not important (since the player may rearrange the cards in his hand any way he chooses); in the problems that follow, we will assume that this is the case unless otherwise stated. Thus we use combinations to compute the possible number of 5-card hands, 52 C 5. This number will go in the denominator of our probability formula, since it is the number of possible outcomes.

For the numerator, we need the number of ways to draw one Ace and four other cards (none of them Aces) from the deck.  Since there are four Aces and we want exactly one of them, there will be 4 C 1 ways to select one Ace; since there are 48 non-Aces and we want 4 of them, there will be 48 C 4 ways to select the four non-Aces.  Now we use the Basic Counting Rule to calculate that there will be 4 C 1 · 48 C 4 ways to choose one ace and four non-Aces.

Putting this all together, we have

[latex]P(\text{oneAce})=\frac{\left({}_{4}{{C}_{1}}\right)\left({}_{48}{{C}_{4}}\right)}{{}_{52}{{C}_{5}}}=\frac{778320}{2598960}\approx0.299[/latex]

Compute the probability of randomly drawing five cards from a deck and getting exactly two Aces.

The solution is similar to the previous example, except now we are choosing 2 Aces out of 4 and 3 non-Aces out of 48; the denominator remains the same:

[latex]P(\text{twoAces})=\frac{\left({}_{4}{{C}_{2}}\right)\left({}_{48}{{C}_{3}}\right)}{{}_{52}{{C}_{5}}}=\frac{103776}{2598960}\approx0.0399[/latex]

It is useful to note that these card problems are remarkably similar to the lottery problems discussed earlier.

View the following for further demonstration of these examples.

Birthday Problem

Let’s take a pause to consider a famous problem in probability theory:

Take a guess at the answer to the above problem. Was your guess fairly low, like around 10%? That seems to be the intuitive answer (30/365, perhaps?). Let’s see if we should listen to our intuition. Let’s start with a simpler problem, however.

Suppose three people are in a room.  What is the probability that there is at least one shared birthday among these three people?

There are a lot of ways there could be at least one shared birthday. Fortunately there is an easier way. We ask ourselves “What is the alternative to having at least one shared birthday?” In this case, the alternative is that there are no shared birthdays. In other words, the alternative to “at least one” is having none . In other words, since this is a complementary event,

P(at least one) = 1 – P(none)

We will start, then, by computing the probability that there is no shared birthday.  Let’s imagine that you are one of these three people.  Your birthday can be anything without conflict, so there are 365 choices out of 365 for your birthday. What is the probability that the second person does not share your birthday?  There are 365 days in the year (let’s ignore leap years) and removing your birthday from contention, there are 364 choices that will guarantee that you do not share a birthday with this person, so the probability that the second person does not share your birthday is 364/365.  Now we move to the third person.  What is the probability that this third person does not have the same birthday as either you or the second person?  There are 363 days that will not duplicate your birthday or the second person’s, so the probability that the third person does not share a birthday with the first two is 363/365.

We want the second person not to share a birthday with you and the third person not to share a birthday with the first two people, so we use the multiplication rule:

[latex]P(\text{nosharedbirthday})=\frac{365}{365}\cdot\frac{364}{365}\cdot\frac{363}{365}\approx0.9918[/latex]

and then subtract from 1 to get

P(shared birthday) = 1 – P(no shared birthday) = 1 – 0.9918 = 0.0082.

This is a pretty small number, so maybe it makes sense that the answer to our original problem will be small.  Let’s make our group a bit bigger.

Suppose five people are in a room.  What is the probability that there is at least one shared birthday among these five people?

Continuing the pattern of the previous example, the answer should be

[latex]P(\text{sharedbirthday})=1-\frac{365}{365}\cdot\frac{364}{365}\cdot\frac{363}{365}\cdot\frac{362}{365}\cdot\frac{361}{365}\approx0.0271[/latex]

Note that we could rewrite this more compactly as

[latex]P(\text{sharedbirthday})=1-\frac{{}_{365}{{P}_{5}}}{{{365}^{5}}}\approx0.0271[/latex]

which makes it a bit easier to type into a calculator or computer, and which suggests a nice formula as we continue to expand the population of our group.

Suppose 30 people are in a room.  What is the probability that there is at least one shared birthday among these 30 people?

Here we can calculate

[latex]P(\text{sharedbirthday})=1-\frac{{}_{365}{{P}_{30}}}{{{365}^{30}}}\approx0.706[/latex]

which gives us the surprising result that when you are in a room with 30 people there is a 70% chance that there will be at least one shared birthday!

The birthday problem is examined in detail in the following.

If you like to bet, and if you can convince 30 people to reveal their birthdays, you might be able to win some money by betting a friend that there will be at least two people with the same birthday in the room anytime you are in a room of 30 or more people. (Of course, you would need to make sure your friend hasn’t studied probability!) You wouldn’t be guaranteed to win, but you should win more than half the time.

This is one of many results in probability theory that is counterintuitive; that is, it goes against our gut instincts.

Suppose 10 people are in a room. What is the probability that there is at least one shared birthday among these 10 people?

  • Revision and Adaptation. Provided by : Lumen Learning. License : CC BY: Attribution
  • Math in Society. Authored by : Lippman, David. Located at : http://www.opentextbookstore.com/mathinsociety/ . License : CC BY: Attribution

Footer Logo Lumen Waymaker

Here is a link to an applet that calculates the answer for permutations and combinations:  Permutation and Combination Calculator . (Note: There are directions on the page that tell you how to enter data and use the applet. There are also example problems; just scroll down the webpage.)

To use the applet : Click on the following link ( Permutation and Combination Calculator ). It will take you to a webpage that looks like this:

where you able to solve problems on permutation and combination

To calculate a permutation:

Let's use the following problem:

The "Visit Europe" travel agency specializes in arranging trips to European cities. On Monday, its first customer, Judy, wanted information on three cities: Rome, Paris, and London. Judy decides that she wants a trip, where she will visit  two  of the three cities. Assuming that she wants to visit each city only one time and  order matters,  how many possible trip arrangements are possible?

To solve this problem using the Combination and Permutation Calculator, do the following:

  • Choose "Count permutations" by clicking on the down arrow of the box next to "Choose goal." 
  • Enter "3" for "Number of sample points in set ( n )", since there are three cities in the set.
  • Enter "2" for "Number of sample points in each permutation ( r )", since we want to have permutations of two cities.
  • Click the "Calculate" button. Note that, for this example, the answer, 6, is displayed in the "Number of permutations ( n  things taken  r at a time)" text box. Thus, there are 6 permutations.

where you able to solve problems on permutation and combination

To calculate a combination:

The "Visit Europe" travel agency specializes in arranging trips to European cities. On Monday, its first customer, Judy, wanted information on three cities: Rome, Paris, and London. Judy decides that she wants a trip, where she will visit two of the cities. Assuming that she wants to visit each city only one time, but order does not matter , how many possible trip arrangements are possible?

  • Choose "Count combinations" by clicking on the down arrow of the box next to "Choose goal." 
  • Enter "3" for "Number of sample points in set (n)" , since there are three cities in the set.
  • Enter "2" for "Number of sample points in each combination ( r )", since we want combinations of 2 cities.
  • Click the "Calculate" button. The answer, 3, is displayed in the "Number of combinations ( n  things taken  r at a time)" text box. Thus, the number of combinations is 3.

where you able to solve problems on permutation and combination

IMAGES

  1. How to solve problems on Permutation and combination very quickly😀(for

    where you able to solve problems on permutation and combination

  2. Permutations P(n,r) (video lessons, examples, solutions)

    where you able to solve problems on permutation and combination

  3. Permutation vs. Combination

    where you able to solve problems on permutation and combination

  4. How to Solve Permutation and Combination problems (part-1b)

    where you able to solve problems on permutation and combination

  5. Permutation and Combination

    where you able to solve problems on permutation and combination

  6. Permutation and combinations: lesson 3 solving problems involving

    where you able to solve problems on permutation and combination

VIDEO

  1. Solving Problems Involving Permutations and Combinations

  2. Permutations and Combinations

  3. Permutation, Combination

  4. permutations and combinations

  5. HOW TO SOLVE PERMUTATION and combinations in Casio fx-991CW

  6. Permutation and Combination

COMMENTS

  1. 12.2: Permutations and Combinations

    Note: The difference between a combination and a permutation is whether order matters or not. If the order of the items is important, use a permutation. If the order of the items is not important, use a combination. Now here are a couple examples where we have to figure out whether it is a permuation or a combination.

  2. Permutations & combinations (practice)

    Permutations & combinations. Google Classroom. You need to put your reindeer, Prancer, Quentin, Rudy, and Jebediah, in a single-file line to pull your sleigh. However, Rudy and Prancer are best friends, so you have to put them next to each other, or they won't fly.

  3. How to Solve Permutations and Combinations? (+FREE Worksheet!)

    Learn how to solve mathematics word problems containing Permutations and Combinations using formulas. There are some examples to help you. ... Step by step guide to solve Permutations and Combinations. Permutations: The number of ways to choose a sample of \(k\) elements from a set of \(n\) distinct objects where order does matter, and ...

  4. Combinations and Permutations

    Combinations. There are also two types of combinations (remember the order does not matter now): Repetition is Allowed: such as coins in your pocket (5,5,5,10,10) No Repetition: such as lottery numbers (2,14,15,27,30,33) 1. Combinations with Repetition. Actually, these are the hardest to explain, so we will come back to this later. 2.

  5. Permutations and Combinations Problems

    This is a combination problem: combining 2 items out of 3 and is written as follows: nC r = n! (n − r)! r! n C r = n! ( n − r)! r! The number of combinations is equal to the number of permutations divided by r! to eliminate those counted more than once because the order is not important. Example 7: Calculate. a) 3C 2 3 C 2.

  6. Permutation vs. Combination

    In math, permutations and combinations are groups or arrangements of things, including people, numbers, and objects. The main difference between the two is that permuations are those groups where ...

  7. 7.7: Probability with Permutations and Combinations

    There are 4 rooms and 5 suspects. This page titled 7.7: Probability with Permutations and Combinations is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

  8. Permutation and Combination Calculator

    Calculate the number of possible combinations. This can be calculated using the combination formula: nCr = n! / (r! (n-r)!) The number of possible combinations, nCr, is 7! / 4! * (7 - 4)! = 35. If the permutations and combinations formula still seems confusing, don't worry; just use our calculator for the calculations.

  9. Counting, permutations, and combinations

    Statistics and probability 16 units · 157 skills. Unit 1 Analyzing categorical data. Unit 2 Displaying and comparing quantitative data. Unit 3 Summarizing quantitative data. Unit 4 Modeling data distributions. Unit 5 Exploring bivariate numerical data. Unit 6 Study design. Unit 7 Probability. Unit 8 Counting, permutations, and combinations.

  10. Permutations, Combinations & Probability (14 Word Problems)

    Learn how to work with permutations, combinations and probability in the 14 word problems we go through in this video by Mario's Math Tutoring. We discuss f...

  11. Easy Permutations and Combinations

    Combination: Choosing 3 desserts from a menu of 10. C (10,3) = 120. Permutation: Listing your 3 favorite desserts, in order, from a menu of 10. P (10,3) = 720. Don't memorize the formulas, understand why they work. Combinations sound simpler than permutations, and they are.

  12. 5.5: Permutations and Combinations

    Unlike permutations, order does not count. Table \(\PageIndex{3}\) is based on Table \(\PageIndex{2}\) but is modified so that repeated combinations are given an "\(x\)" instead of a number. For example, "yellow then red" has an "\(x\)" because the combination of red and yellow was already included as choice number \(1\). As you can see, there ...

  13. Permutation vs Combination: Differences & Examples

    Permutations: The order of outcomes matters. Combinations: The order does not matter. Let's understand this difference between permutation vs combination in greater detail. And then you'll learn how to calculate the total number of each. In some scenarios, the order of outcomes matters. For example, if you have a lock where you need to ...

  14. 15 Examples Of Permutations And Combinations

    Example 5) Find out the number of ways a judge can award a first, second, and third place in a contest with 18 competitors. solution: Here n=18, r=3. so Required number of ways=. nPr = n !/ ( n-r )! 18P3 = 18!/ (18-3)! = 15!.16.17.18/15! = 4896. Among the 18 contestants, in 4896 number of ways, a judge can award a 1st, 2nd and 3rd place in a ...

  15. Permutations and Combinations Questions (With Solutions)

    What are permutations and combinations? A permutation is an arrangement in a definite order of a number of objects taken, some or all at a time. The combination is a way of selecting items from a collection, such that (unlike permutations) the order of selection does not matter. Also, read: Permutation and combination. Permutations and ...

  16. Permutation And Combination Solved Problems

    Defining Permutations and Combinations. The terms permutations and combinations revolve around the concept of arranging and selecting objects from a given set. When the sequence matters, we refer to it as a permutation. On the other hand, when the arrangement doesn't matter, it's called a combination.

  17. How to Solve Difficult Permutations & Combination ...

    Payal Tandon, GMAT Club's top-rated instructor, outlines a unified approach to solve 700-level permutation and combination and probability questions.

  18. Permutations

    Summary of permutations. A permutation is a list of objects, in which the order is important. Permutations are used when we are counting without replacing objects and order does matter. If the order doesn't matter, we use combinations. In general P(n, k) means the number of permutations of n objects from which we take k objects. Alternatively ...

  19. Permutation and Combination

    Combinations are used when the same kind of things are to. be sorted. Permutation of two things out of three given things. a, b, c is ab, ba, bc, cb, ac, ca. the combination of two things from three given things. a, b, c is ab, bc, ca. Formula for permuation is: n Pr = n!/ (n - r)!

  20. Permutations and combinations

    permutations and combinations, the various ways in which objects from a set may be selected, generally without replacement, to form subsets. This selection of subsets is called a permutation when the order of selection is a factor, a combination when order is not a factor. By considering the ratio of the number of desired subsets to the number ...

  21. Probability Using Permutations and Combinations

    Example. In the state lottery from the previous example, if five of the six numbers drawn match the numbers that a player has chosen, the player wins a second prize of $1,000. Compute the probability that you win the second prize if you purchase a single lottery ticket. Show Solution. The previous examples are worked in the following video.

  22. How do I determine whether to use permutations or combinations to solve

    For combinations, you must also divide out the "repetitions," so to say, i.e., divide by $3!$ in this case. Notice the answer is smaller. Other common permutation problems include making passwords and ordering books on a shelf, while combinations are tackled in poker hand problems and committee problems (in which each member has a specific ...

  23. Permutation and Combination Applet

    To solve this problem using the Combination and Permutation Calculator, do the following: Choose "Count permutations" by clicking on the down arrow of the box next to "Choose goal." Enter "3" for "Number of sample points in set ( n )", since there are three cities in the set. Enter "2" for "Number of sample points in each permutation ( r ...