Permutations Worksheets

What are Permutations? The words permutations and combinations are most commonly used together whenever they are used anywhere in mathematics. Before you know what alterations are, lets' first try to understand what a combination is. Consider this example to understand it; Mother made a salad mix of walnut, lettuce, cherry tomatoes, and apple. Here, in this example, it does not matter what order you follow to mix the ingredients. You dump in all the veggies and fruits and prepare a bowl of salad for yourself. This salad is just a combination of fruits and vegetables without any order. If the order doesn’t matter, then it is known as a combination, and if the order matters, then it is called a permutation. Therefore, you can say that permutation is an ordered combination. A permutation is an ordered arrangement of all or a few parts of a set of objects, ensuring that the arrangement does not lose.

Basic Lesson

Demonstrates how to evaluate simple permutations.

Intermediate Lesson

Explores how to work with word problem based permutations. In how many ways can 4 players be listed in a game?

Independent Practice 1

Contains 20 Permutation problems. The answers can be found below.

Independent Practice 2

Example: How many different 5-letter arrangements are there of the letters in the word PLANT?

Homework Worksheet

Permutation problems for students to work on at home. Example problems are provided and explained.

We work on a wide range of permutation problems. A math scoring matrix is included.

Homework and Quiz Answer Key

Answers for the homework and quiz.

Lesson and Practice Answer Key

Answers for both lessons and both practice sheets.

Demonstrates how to use the permutation equation to determine the possible number of outcomes for an event. A permutation is a selection of objects in which the order of the objects matters. A formula for the number of possible permutations of k objects from a set of n.

Explains how to determine total outcomes of an event in short form. Practice problems are provided. Two cards are drawn at random from a deck of 52 cards, without replacement. How many ways we can draw these 2 cards, 7 and a king (in that order)? There are 52 cards.

Austin, John, and Brittany ran in a race. In how many different orders can they finish the race? The answers can be found below.

Example: How many permutations can you make from the letters A through D?

12 permutations problems for students to work on at home. Example problems are provided and explained.

10 permutations problems. A math scoring matrix is included.

Mary Everest Boole, Philosophy And Fun Of Algebra

"let us agree to call it x, and let us always remember that x stands for the Unknown. ... This method of solving problems by honest confession of one's ignorance is called Algebra."

Did you know that famed mathematician Rene Descartes was admitted into college at the tender age of only eight?!

Print book

Permutations & Combinations

Description, table of contents, fundamental counting principal, discovery answer key, video lessons, video lesson, interactive activity.

Guided Practice

When trying to determine the total number of possible outcomes, it is sometimes hard to create the entire sample space either by organized list or by tree diagram. When this is true, it is helpful to use the fundamental counting principle. The fundamental counting principle states that when the outcomes of two or more events are independent, the total number of outcomes can be determined by multiplying the number of outcomes for each individual event.

Example A sandwich shop offers 5 kinds of bread, 8 types of meat and 6 cheeses. If you choose to make a sandwich with one item from each category, how many total possible sandwiches can you create?

Multiply the total outcomes from each category.

There are 240 total possible sandwiches.

To learn more about the fundamental counting principle, select the following link:

Permutations

A permutation is an arrangement of objects in a certain order. The word arrangement is used, if the order of things is considered . An example of a permutation is selecting three students to be President, Vice President, and Secretary of the student council. The order in which they are selected will change the job they are given.

Permutations require the use of factorials, which are represented by the symbol "!". See the following example on how to apply the factorial symbol.

Factorial Example Find 6!

6! = 6?5?4?3?2?1 = 720

To do an activity that will help you develop the formula for permutations, select the following link:

Permutations Formula Worksheet

*Note: If Google Docs displays, “Sorry, we were unable to retrieve the document for viewing,” refresh your browser.

Once you have worked through the worksheet, select the following link to check your understanding:

Permutations Formula Answer Key

How many different 7 player batting orders can be created from a 9 player team?

Step 1. Determine if this is a combination or a permutation.

Since the order the batters are in makes a difference, this is a permutation.

Step 2. Enter the data into the formula.

Perm_ex1

Step 3. Simplify the formula.

Perm_ex2

There are 181,440 different batting orders to choose from.

To learn more about permutations, select the following link to a Holt, Rinehart & Winston video lesson:

Finding Permutations

To learn even more about permutations, watch the following video:

Combinations

A combination is a selection of objects in no particular order. The word selection is used, when the order of things has no importance . An example of a combination is selecting three students to be on the student council. Since no specific jobs are given, the order they are chosen will not change the members of the council.

To do an experiment that will help you develop the formula for combinations, select the following link:

Combinations Formula Worksheet

Once you have worked through the worksheet, select the following link to check your understanding: Combinations Formula Answer Key

How many different 5 player basketball teams can be selected from an 11 player team?

Since the order of the players does not make a difference, this is a combination.

Combex1

There are 462 different teams of 5 players to choose from.

Combinations & Permutations Calculator

Combination vs. Permutation

To summarize, permutations are used when the order does matter and combinations are used when the order does not matter. The following formulas can be used to determine permutations and combinations:

Perm vs Comb

To learn more about combinations vs. permutations, select the following link:

Combinations & Permutations

To learn how to use your TI-84 calculator to calculate combinations and permutations, watch the following video:

To solidify your understanding of permutations and combinations, visit the following link to Holt, Rinehart, and Winston Homework Help Online. It provides examples, video tutorials, and interactive practice with answers available. The Practice and Problem Solving section has two parts. The first part offers practice with a complete video explanation for the type of problem with just a click of the video icon. The second part offers practice with the solution for each problem only a click of the light bulb away.

Brightstorm.com, "Combinations and Permutations Video." http://www.brightstorm.com/math/algebra-2/combinatorics/combinations-vs-permutations (accessed 7/27/2010).

Embracing Mathematics, Assessment & Technology in High Schools; a Michigan Mathematics & Science Partnership Grant Project

Holt, Rinehart & Winston, "Finding Permutations." http://my.hrw.com/math06_07/nsmedia/lesson_videos/alg2/player.ht ml?contentSrc=7199/7199.xml (accessed 7/28/2010).

Holt, Rinehart & Winston, "Probability & Statistics." http://my.hrw.com/math06_07/nsmedia/homework_help/alg2/alg2_ch 11_01_homeworkhelp.html (accessed 7/28/2010).

Holt, Rinehart & Winston, "Solving Combination Problems." http://my.hrw.com/math06_07/nsmedia/lesson_videos/alg2/player.ht ml?contentSrc=7200/7200.xml (accessed 7/28/2010).

Holt, Rinehart & Winston, "Using the Fundamental Counting Principal." http://my.hrw.com/math06_07/nsmedia/lesson_videos/alg2/player.ht ml?contentSrc=6490/6490.xml (accessed 7/28/2010).

Mathisfun.com, "Combinations and Permutations Calculator." http://www.mathsisfun.com/combinatorics/combinations-permutations- calculator.html (accessed 7/28/2010).

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Mathematics LibreTexts

12.2: Permutations and Combinations

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  • Page ID 92814

  • Maxie Inigo, Jennifer Jameson, Kathryn Kozak, Maya Lanzetta, & Kim Sonier
  • Coconino Community College

Consider the following counting problems:

  • In how many ways can three runners finish a race?
  • In how many ways can a group of three people be chosen to work on a project?

What is the difference between these two problems? In the first problem the order that the runners finish the race matters. In the second problem the order in which the three people are chosen is not important, only which three people are chosen matters.

Permutation

A permutation is an arrangement of a set of items. The number of permutations of n items taking r at a time is given by:​​​

\[P(n, r)=\frac{n !}{(n-r) !} \label{permutation}\]

Note: Many calculators can calculate permutations directly. Look for a function that looks like \(_nP_r\) or \(P(n,r)\)

Let’s look at a simple example to understand the formula for the number of permutations of a set of objects. Assume that 10 cars are in a race. In how many ways can three cars finish in first, second and third place? The order in which the cars finish is important. Use the multiplication principle. There are 10 possible cars to finish first. Once a car has finished first, there are nine cars to finish second. After the second car is finished, any of the eight remaining cars can finish third. 10 x 9 x 8 = 720. This is a permutation of 10 items taking three at a time.

Using the permutation formula:

\[P(10,3)=\frac{10 !}{(10-3) !}=\frac{10 !}{7 !}= \dfrac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = 10 \cdot 9 \cdot 8 = 720 \nonumber\]

Using the fundamental counting principle:

\[\underline{10} \cdot \underline{9} \cdot \underline{8} = 720 \nonumber\]

There are 720 different ways for cars to finish in the top three places.

The school orchestra is planning to play six pieces of music at their next concert. How many different programs are possible?

This is a permutation because they are arranging the songs in order to make the program. Using the permutation formula:

\[P(6,6)=\frac{6 !}{(6-6) !}=\frac{6 !}{0 !}=\frac{720}{1}=720 \nonumber\]

\[\underline{6} \cdot \underline{5} \cdot \underline{4} \cdot \underline{3} \cdot \underline{2} \cdot \underline{1}=720 \nonumber\]

There are 720 different ways of arranging the songs to make the program.

The Volunteer Club has 18 members. An election is held to choose a president, vice-president and secretary. In how many ways can the three officers be chosen?

The order in which the officers are chosen matters so this is a permutation.

\[P(18,3)=\frac{18 !}{(18-3) !}=\frac{18 !}{15 !}=18 \cdot 17 \cdot 16=4896 \nonumber\]

Note: All digits in 18! in the numerator from 15 down to one will cancel with the 15! in the denominator.

Using the fundamental principle:

\(\begin{array} {cccccc} {\underline{18}}&{\cdot}&{\underline{17}}&{\cdot}&{\underline{16}}&{ = 4896}\\ {\text{Pres.}}&{}&{\text{V.P.}}&{}&{\text{Sec.}}&{} \end{array}\)

There are 4896 different ways the three officers can be chosen.

Another notation for permutations is \(_nP_r\). So, \(P(18,3)\) can also be written as \(_{18}P_3\). Most scientific calculators have an \(_nP_r\) button or function.

Combinations are when the order does not even matter. We are just collecting objects together.

Choose a committee of two people from persons A, B, C, D and E. By the multiplication principle there are \(5 \cdot 4 = 20\) ways to arrange the two people.

AB AC AD AE BA BC BD BE CA CB

CD CE DA DB DC DE EA EB EC ED

Committees AB and BA are the same committee. Similarly for committees CD and DC. Every committee is counted twice.

\[\frac{20}{2}=10 \nonumber\]

so there are 10 possible different committees.

Now choose a committee of three people from persons A, B, C, D and E. There are \(5 \cdot 4 \cdot 3 = 60\) ways to pick three people in order. Think about the committees with persons A, B and C. There are \(3! =6\) of them.

ABC ACB BAC BCA CAB CBA

Each of these is counted as one of the 60 possibilities but they are the same committee. Each committee is counted six times so there are

\[\frac{60}{6}=10 \, \text{different committees}. \nonumber\]

In both cases we divided the number of permutations by the number of ways to rearrange the people chosen.

The number of permutations of n people taking r at a time is \(P(n,r)\) and the number of ways to rearrange the people chosen is \(r!\). Putting these together we get

\[\begin{aligned} \frac{n !}{\# \text { ways to arrange r items }} &=\frac{P(n, r)}{r !}=\frac{(n-r) !}{\frac{r !}{1}} \\ &=\frac{n !}{(n-r) !} \cdot \frac{1}{r !} \\ &=\frac{n !}{(n-r) ! r !} \end{aligned}\]

Combination

A combination is a selection of objects in which the order of selection does not matter. The number of combinations of n items taking r at a time is:

\[C(n, r)=\frac{n !}{r !(n-r) !} \label{combination}\]

Note: Many calculators can calculate combinations directly. Look for a function that looks like \(_nC_r\) or \(C(n,r)\) .

A student has a summer reading list of eight books. The student must read five of the books before the end of the summer. In how many ways can the student read five of the eight books?

The order of the books is not important, only which books are read. This is a combination of eight items taking five at a time.

\[C(8,5)=\frac{8 !}{5 !(8-5) !}=\frac{8 !}{5 ! 3 !}= \dfrac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{5 \times 4 \times 3 \times 2 \times 1 \times 3 \times 2 \times 1} = \dfrac{8 \times 7 \times 6}{3 \times 2 \times 1} = 8 \times 7 = 56\]

There are 56 ways to choose five of the books to read.

A child wants to pick three pieces of Halloween candy to take in her school lunch box. If there are 13 pieces of candy to choose from, how many ways can she pick the three pieces?

This is a combination because it does not matter in what order the candy is chosen.

\[\begin{align*} C(13,3) &=\frac{13 !}{3 !(13-3) !}=\frac{13 !}{3 ! 10 !} \\[4pt] &= \dfrac{13 \times 12 \times 11 \times 10 \times 9\times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1)(10 \times 9\times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)} \\[4pt] &=\frac{13 \times 12 \times 11}{3 \times 2 \times 1} \\[4pt] =\frac{1716}{6}=286 \end{align*}\]

There are 286 ways to choose the three pieces of candy to pack in her lunch.

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.

A serial number for a particular model of bicycle consists of a letter followed by four digits and ends with two letters. Neither letters nor numbers can be repeated. How many different serial numbers are possible?

This is a permutation because the order matters.

Use the multiplication principle to solve this. There are 26 letters and 10 digits possible.

\[26 \cdot 10 \cdot \underline{9} \cdot \underline{8} \cdot \underline{7} \cdot \underline{25} \cdot \underline{24}=78,624,000\]

There are 78,624,000 different serial numbers of this form.

A class consists of 15 men and 12 women. In how many ways can two men and two women be chosen to participate in an in-class activity?

This is a combination since the order in which the people is chosen is not important.

Choose two men:

\[C(15,2)=\frac{15 !}{2 !(15-2) !}=\frac{15 !}{2 ! 13 !}=105 \nonumber\]

Choose two women:

\[C(12,2)=\frac{12 !}{2 !(12-2) !}=\frac{12 !}{2 ! 10 !}=66 \nonumber\]

We want 2 men and 2 women so multiply these results.

\[105(66)=6930 \nonumber\]

There are 6930 ways to choose two men and two women to participate.

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6th grade (Illustrative Mathematics)

Unit 1: area and surface area, unit 2: introducing ratios, unit 3: unit rates and percentages, unit 4: dividing fractions, unit 5: arithmetic in base ten, unit 6: expressions and equations, unit 7: rational numbers, unit 8: data sets and distribution.

PERMUTATION PRACTICE PROBLEMS WITH ANSWERS

Permutation Practice Problems with Answers :

Here we are going to see some practice questions based on the concept permutation.

Permutation Practice Problems with Answers

Problem 1 : 

Determine the number of permutations of the letters of the word SIMPLE if all are taken at a time?

Number of letters in the word "SIMPLE"  =  6

All are unique letters.

Number of permutation  = 6 P 6   =  6!

  =  6  ⋅ 5  ⋅ 4  ⋅ 3  ⋅ 2  ⋅ 1

  =  720

Hence total number of permutation is 720.

Problem 2 :

A test consists of 10 multiple choice questions. In how many ways can the test be answered if

(i) Each question has four choices?

(ii) The first four questions have three choices and the remaining have five choices?

(iii) Question number n has n + 1 choices?

Number of ways to answer 1 st question  =  4

Number of ways to answer 2 nd  question  =  4

Number of ways to answer 3 rd  question  =  4

............................

Number of ways  =  4  ⋅ 4  ⋅ 4  ⋅ 4  ⋅ 4  ⋅ 4  ⋅ 4  ⋅ 4  ⋅ 4  ⋅ 4

  =  4 10

Hence the total number of ways  =   4 10

Number of ways to answer 1 st question  =  3

Number of ways to answer 2 nd  question  =  3

Number of ways to answer 3 rd  question  =  3

Number of ways to answer 4 th  question  =  3

Number of ways to answer 5 th  question  =  5

Number of ways to answer 6 th  question  =  5

.............................. 

Number of ways  =    3  ⋅ 3  ⋅ 3  ⋅ 3  ⋅ 5  ⋅ 5   ⋅ 5  ⋅ 5  ⋅ 5  ⋅ 5

  =  3 4 ⋅ 5 6

Hence the total number of ways is  3 4 ⋅ 5 6 .

Number of ways to answer 1 st question  =  2

Number of ways to answer 2 nd question  =  3

Number of ways to answer 4 th  question  =  5

...................

Number of ways to answer 10 th  question  =  11

Number of ways  =  1  ⋅  2  ⋅  3  ⋅  4  ⋅  5  ⋅  6  ⋅  7  ⋅  8  ⋅ 9  ⋅  10  ⋅  11

  =  11!

Hence the total number of ways is 11!.

lesson 6 problem solving practice permutations answer key

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lesson 6 problem solving practice permutations answer key

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?

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COMMENTS

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