Prepare the **important questions for class 11 Maths Chapter 7 – Permutations and Combinations, **which is given here. Students can learn how to find the factorial of a number, and how to compute the permutations and combinations for the given condition. This chapter concerns with computing the number of different ways of the arrangement of a given number of objects without listing the objects. Also, prepare all the chapters important questions for class 11 Maths here.

Class 11 Maths Chapter 7 – permutations and combinations describe the following important concepts:

- Fundamental Principle of Counting
- Multiplication Principle
- Addition Principle

- Permutations
- Permutation of n different objects
- When the repetition of an object is allowed
- Permutations when the objects are not distinct

- Combinations

**Also, Check: **

- Important 1 Mark Questions for CBSE Class 11 Maths
- Important 4 Marks Questions for CBSE Class 11 Maths
- Important 6 Marks Questions for CBSE Class 11 Maths

## Class 11 Chapter 7 – Permutations and Combinations Important Questions with Solutions

Practice the given important questions for class 11 Maths Chapter 7 – permutations and combinations given below.

**Question 1: **

Find the 3-digit numbers that can be formed from the given digits: 1, 2, 3, 4 and 5 assuming that

**a)** digits can be repeated.

**b)** digits are not allowed to be repeated.

**Solution:**

**a)** By the multiplication principle, the number of ways in which three-digit numbers can be formed from the given digits is 5 Ã— 5 Ã— 5 = 125

**b)** By the multiplication principle, the number of ways in which three-digit numbers can be formed without repeating the given digits is 5 Ã— 4 Ã— 3 = 60

**Question 2: **

A coin is tossed 6 times, and the outcomes are noted. How many possible outcomes can be there?

**Solution: **

When we toss a coin once, the number of outcomes we get is 2 (Either Head or tail)

So, in each throw, the no. of ways to get a different face will be 2.

Therefore, by the multiplication principle, the required no. of possible outcomes is

2 x 2 x 2 x 2 Ã— 2 Ã— 2 = 64

**Question 3:**

Evaluate the following

(i) 6 ! (ii) 5 ! â€“ 2 !

**Solution:**

**(i)** 6! = 1 Ã— 2 Ã— 3 Ã— 4 Ã— 5 Ã— 6 = 720

**(ii)** 5! = 1 Ã— 2 Ã— 3 Ã— 4 x 5 = 120

As 2! = 1 Ã— 2 = 2

Therefore,Â 5 ! â€“ 2 ! = 120-2 = 118.

**Question 4:**

From a team of 6 students, in how many ways can we choose a captain and vice-captain assuming one person can not hold more than one position?

**Solution:**

From a team of 6 students, two students are to be chosen in such a way that one student will hold only one position.

Here, the no. of ways of choosing a captain and vice-captain is the permutation of 6 different things taken 2 at a time.

So, ^{6}P_{2} = 6! / ( 6 -2 )! = 6! / 4! = 30

**Question 5:**

How many words, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter Exactly once?

**Solution:**

Number of letters in word EQUATION` = 8

n = 8

If all letters of the word used at a time

r = 8

Different numbers formed = nPr

= ^{8}P_{8}

= 8!/(8 8)!

= 8!/0!

= 8!/1

= 8!

= 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1

= 40320

**Question 6:**

How many words can be formed each of 2 vowels and 3 consonants from the letters of the given word – DAUGHTER?

**Solution:**

No. of Vowels in the word – DAUGHTER is 3.

No. of Consonants in the word Daughter is 5.

No of ways to select a vowel = ^{3}c_{2} = 3!/2!(3 – 2)! = 3

No. of ways to select a consonant = ^{5}c_{3 }= 5!/3!(5 – 3)! = 10

Now you know that the number of combinations of 3 consonants and 2 vowels = 10x 3 = 30

Total number of words = 30 x 5! = 3600 ways.

**Question 7:**

It is needed to seat 5 boys and 4 girls in a row so that the girl gets the even places. How many are such arrangements possible?

**Solution:**

5 boys and 4 girls are to be seated in a row so that the girl gets the even places.

The 5 boys can be seated in 5! Ways.

For each of the arrangement, the 4 girls can be seated only at the places which are cross marked to make girls occupy the even places).

B x B x B x B x B

So, the girls can be seated in 4! Ways.

Hence, the possible number of arrangements = 4! Ã— 5! = 24 Ã— 120 = 2880

**Question 8:**

Find the number of 5-card combinations out of a deck of 52 cards if each selection of 5 cards has exactly one king.

**Solution:**

Take a deck of 52 cards,

To get exactly one king, 5-card combinations have to be made. It should be made in such a way that in each selection of 5 cards, or in a deck of 52 cards, there will be 4 kings.

To select 1 king out of 4 kings = ^{4}c_{1}

To select 4 cards out of the remaining 48 cards = ^{48}c_{4}

To get the needed number of 5 card combination = ^{4}c_{1} x ^{48}c_{4 }

= 4x2x 47x 46×45

= 778320 ways.

**Question 9:**

Find the number of 6 digit numbers that can be formed by using the digits 0, 1, 3, 5, 7, and 9. These digits shall be divisible by 10, and no digit shall be repeated?

**Solution:**

The number which has a 0 in its units place is divisible by 10.

If we put 0 in the unit place, _ _ _ _ 0, there will be as many ways to fill 5 vacant places. (1, 3, 5, 7, 9)

The five vacant places can be filled in 5! ways = 120.

**Question 10:**

Evaluate: 10! – 6!

**Solution:**

10! = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x1 = 3628800

6! = 6 X 5 x 4 x 3 x 2 x 1 = 720

10! â€“ 6! = 3628800 – 720 = 3628080

### Practice Problems for Class 11 Maths Chapter 7

Solve the problems from class 11 Maths Chapter 7 – permutations and combinations:

- Calculate how many numbers are there between 99 and 1000 having atleast one of their digits 7?
- A boy has 3 library tickets and 8 books of his interest in the library. Of these 8, he does not want to borrow Mathematics Part II, unless Mathematics Part I is also borrowed. In how many ways can he choose the three books to be borrowed?
- Given that 5 flags are of different colours. Calculate how many different signals can be generated if each signal requires the use of 2 flags, one below the other?
- Compute n!/ r!(n-r)!, when n=5 and r=2.
- Determine the number of permutations of the letters of the word ALLAHABAD.

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GOOD QUESTION

nice

Really nice questions helped a lot in practice!!