Class 11 Maths Chapter 7 Permutations and Combinations MCQs

Class 11 Maths Chapter 7 Permutations and Combinations MCQs are made available online to assist students in achieving good exam results. The multiple-choice questions follow the CBSE curriculum (2022-2023) and NCERT norms. Answers and full explanations are provided for these multiple-choice questions. Click here to access all chapter-by-chapter MCQs for Class 11 Maths.

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MCQs on Class 11 Maths Chapter 7 Permutations and Combinations

Check out the MCQs for Class 11 Maths Chapter 7 Permutations and Combinations. There are four options for each MCQ, but only one of them is accurate. Students should select the proper option and double-check their answers against the solutions on our page. Also, make sure to look at the important questions for class 11 Maths.

1) The value of P(n, n-1) is

1. n
2. n!
3. 2n
4. 2n!

Answer: (b) n!

Explanation: We know that P(n, r) = ⁿP_r = n!/(n-r)!

Hence, P(n, n-1) = ⁿP_n-1 = n!/[n-(n-1)]!

P(n, n-1) = n!/(n-n+1)! = n!/1! = n!

Therefore, the value of P(n, n-1) is n!.

Hence, the correct answer is option (b) n!

2) The number of ways in which 8 students can be seated in a line is

1. 5040
2. 50400
3. 40230
4. 40320

Answer: (d) 40320

Explanation: For the 1st position, there are 8 possible choices. For the 2nd position, there are 7 possible choices. For the 3rd position, there are 7 possible choices, etc. And for the eighth position, there is only one possible choice. Hence, this can be written as 8!

(i.e.) 8! = 8×7×6×5×4×3×2×1=40,320

Hence, the number of ways in which 8 students can be seated in a line is 40320.

3) If ⁿP₅ = 60ⁿ⁻¹P₃, the value of n is

1. 6
2. 10
3. 12
4. 16

Answer: (b) 10

Explanation:

Given that ⁿP₅ = 60ⁿ⁻¹P₃

We know that P(n, r) = ⁿP_r = n!/(n-r)!

Now, apply the formula on both sides to get the value of n.

n!/(n-5)! = 60 [(n-1)!/[(n-1)-3]!]

On solving the above equation, we get n= -6 and n=10.

Since the value of n cannot be negative, the value of n is 10.

Hence, option (b) 10 is the correct answer.

4) The number of squares that can be formed on a chessboard is

1. 64
2. 160
3. 204
4. 224

Answer: (c) 204

Explanation:

1×1 grid squares = 8×8 = 64, 2×2 grid squares = 7×7 = 49, 3×3 grid squares = 6×6 = 36 upto 8×8 grid squares = 1×1 = 1.

Hence, the total number of squares that can be formed on a chess board = 8²+7²+6²+…+1²

= 1²+2²+3²+…+8²

= [n(n+1)(2n+1)]/6

Here, n=8

Hence,

= [8(8+1)(16+1)]/6

= (8×9×17)/6

= 12×17 = 204

Hence, option (c) 204 is the correct answer.

5) The number of ways 4 boys and 3 girls can be seated in a row so that they are alternate is

1. 12
2. 104
3. 144
4. 256

Answer: (c) 144

Explanation: Given that, there are 4 boys and 3 girls.

The only pattern 4 boys and 3 girls are arranged in an alternate way is BGBGBGB.

Therefore, the total number of ways is 4!×3! = 144.

6) The number of ways 10 digit numbers can be written using the digits 1 and 2 is

1. 2¹⁰
2. ¹⁰C₂
3. 10!
4. ¹⁰C₁+⁹C₂

Answer: (a) 2¹⁰

Explanation: Given digits are 1 and 2.

Here, each place can be filled in two ways either with 1 or 2 and every place has two chances.

Therefore, the number of ways 10 digit numbers can be written using the digits 1 and 2 is 2¹⁰.

Hence, option (a) 2¹⁰ is the correct answer.

7) A coin is tossed n times, the number of all the possible outcomes is

1. 2n
2. 2ⁿ
3. C(n, 2)
4. P(n, 2)

Answer: (b) 2ⁿ

Explanation:

We know that, when a coin is tossed, we will get either head or tail.

Therefore, the number of all possible outcomes when a coin is tossed n times is 2ⁿ.

8) In how many ways 8 distinct toys can be distributed among 5 children?

1. ⁸P₅
2. ⁵P₈
3. 5⁸
4. 8⁵

Answer: (c) 5⁸

Explanation:

Given that, the number of toys = 8

The number of children = 5.

Hence, the number of ways 8 distinct toys can be distributed among 5 children is 5×5×5×5×5×5×5×5=5⁸.

Hence, option (c) 5⁸ is the correct answer.

9) There are 10 true-false questions in an examination. These questions can be answered in:

1. 20 ways
2. 100 ways
3. 512 ways
4. 1024 ways

Answer: (d) 1024 ways

Explanation:

Given that there are 10 questions.

Each question can be answered in two ways. (i.e. either true or false).

Hence, the number of ways these questions can be answered is 2¹⁰, which is equal to 1024.

10) In how many ways can we paint the six faces of a cube with six different colours?

1. 30
2. 6
3. 6!
4. None of the above

Answer: (a) 30

Explanation:

Now, let us consider the 6 different colours: c1, c2, c3, c4, c5, c6.

Assume that the face of the cube facing up is c1. So, the face of the cube at the bottom can be painted in 5 different ways.

So, 4 faces on the horizontal side of the cube are in circular permutation and they can be painted in (4-1)! ways.

Hence, the total number of ways we can paint the faces of a cube with six different colours is 5×(4-1)! ways.

Hence, 5×(4-1)! ways = 5×3! = 5×3×2×1 = 30 ways.

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