CBSE Important Questions Class 11 Maths Chapter 6 – Permutations and Combinations

Class 11 Maths Chapter 6 Permutations and Combinations is part of the CBSE Class 11 Maths syllabus. It explains the fundamental principle of counting and then builds on it to define factorials, permutations (where order matters) and combinations (where order does not matter). Students learn the formulae nPr and nCr and use them to count arrangements of letters, digits, committees and teams.

These CBSE important questions for Class 11 Maths Chapter 6 cover seven-digit numbers divisible by 9, assertion-reason questions on choosing ice creams and forming a committee, arranging the letters of a word, forming numbers with and without repetition, and the difference between a permutation and a combination. Every question has a clear step-by-step solution in simple English, so students can practise on their own, parents can check the working, and teachers can use them for class tests.

Q1. The number of seven-digit numbers divisible by 9 formed with the digits 1, 2, 3, 4, 5, 6, 7, 8, 9 without repetition is
(a) 7!
(b) ⁹P₇
(c) 3(7!)
(d) 4(7!)

Answer: (d) 4(7!)
The sum of all nine digits 1 to 9 is 45, which is divisible by 9.
To make a seven-digit number we leave out two digits. For the remaining sum to stay divisible by 9, the two omitted digits must themselves add up to 9.
Such pairs are (1,8), (2,7), (3,6) and (4,5) — that is 4 pairs.
For each pair, the remaining 7 digits can be arranged in 7! ways.
Total = 4 × 7!

Q2. Read the statements and select the correct option.
Assertion (A): The number of ways to choose 2 ice creams from 10 different flavours is 90.
Reason (R): The order of selection of ice creams does not matter when choosing 2.
(a) Both A and R are true, and R is the correct explanation of A
(b) Both A and R are true, but R is not the correct explanation of A
(c) A is true, but R is false
(d) A is false, but R is true

Answer: (d) A is false, but R is true
Since order does not matter, we use combinations:
¹⁰C₂ = (10 × 9) / (2 × 1) = 45, not 90.
The number 90 would be ¹⁰P₂, which counts the order as well. So the Assertion is wrong while the Reason is correct.

Q3. Read the statements and select the correct option.
Assertion (A): When choosing a committee of 4 from 7 people, the number of ways is ⁷C₄.
Reason (R): The order of selection of members does not matter in a committee.
(a) Both A and R are true, and R is the correct explanation of A
(b) Both A and R are true, but R is not the correct explanation of A
(c) A is true, but R is false
(d) A is false, but R is true

Answer: (a) Both A and R are true, and R is the correct explanation of A
In a committee, only who is selected matters, not the order in which they are chosen. That is exactly the situation where combinations are used, so the number of ways is ⁷C₄ = 35. The Reason gives the reason for using C rather than P.

Q4. Write the difference between a permutation and a combination. Give one example of each.

Answer: A permutation is an arrangement in which the order matters. Its formula is
nPr = n! / (n − r)!
Example: arranging 3 books out of 5 on a shelf, because changing the order gives a different arrangement.
A combination is a selection in which the order does not matter. Its formula is
nCr = n! / [r!(n − r)!]
Example: choosing 3 students out of 5 for a team, because the same three students form the same team in any order.
The two are related by nPr = nCr × r!

Q5. How many words, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter exactly once?

Answer: The word EQUATION has 8 letters, and all of them are different.
The number of arrangements of 8 distinct letters is 8!.
8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
= 40320 words

Q6. How many 4-digit numbers can be formed with the digits 1, 2, 3, 4, 5 if (i) repetition is allowed and (ii) repetition is not allowed?

Answer: (i) With repetition: Each of the four places can be filled by any of the 5 digits.
Total = 5 × 5 × 5 × 5 = 5⁴ = 625
(ii) Without repetition: The first place has 5 choices, the second 4, the third 3 and the fourth 2.
Total = 5 × 4 × 3 × 2 = 120
(This is the same as ⁵P₄ = 5!/1! = 120.)

Q7. Find the number of arrangements of the letters of the word INDEPENDENCE. In how many of these do all the vowels always occur together?

Answer: INDEPENDENCE has 12 letters, with N appearing 3 times, D twice, E four times, and I, P, C once each.
Total arrangements = 12! / (3! × 2! × 4!)
= 479001600 / (6 × 2 × 24)
= 1663200
Vowels together: The vowels are I, E, E, E, E — five of them. Treat them as one block.
Then we arrange 8 units (the block plus N, D, P, N, D, N, C), where N repeats 3 times and D twice:
8! / (3! × 2!) = 40320/12 = 3360
The vowels inside the block can be arranged in 5!/4! = 5 ways.
Total = 3360 × 5 = 16800

Q8. In how many ways can a committee of 3 girls and 4 boys be chosen from 7 girls and 7 boys?

Answer: Since this is a selection, we use combinations.
Choosing 3 girls from 7: ⁷C₃ = (7 × 6 × 5)/(3 × 2 × 1) = 35
Choosing 4 boys from 7: ⁷C₄ = (7 × 6 × 5 × 4)/(4 × 3 × 2 × 1) = 35
By the fundamental principle of counting, the two choices are made together:
Total = 35 × 35 = 1225 ways

Q9. Evaluate 8! and 8! − 6!, and simplify 10!/(7! × 3!).

Answer: 8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 40320
6! = 720, so 8! − 6! = 40320 − 720 = 39600
10!/(7! × 3!): Write 10! as 10 × 9 × 8 × 7!, so the 7! cancels:
= (10 × 9 × 8) / (3 × 2 × 1)
= 720/6
= 120
(This is ¹⁰C₃.)

Q10. If nC₂ = 15, find the value of n.

Answer: nC₂ = n(n − 1) / 2
So n(n − 1)/2 = 15
n(n − 1) = 30
n² − n − 30 = 0
(n − 6)(n + 5) = 0
So n = 6 or n = −5. Since n must be a positive whole number, n = 6.

Q11. In how many ways can 5 people be seated around a circular table?

Answer: In a circular arrangement, what matters is the position of each person relative to the others, not the absolute seat. Rotating everyone by one seat gives the same arrangement.
So we fix one person's place and arrange the remaining (n − 1) people.
Number of circular arrangements = (n − 1)!
Here n = 5, so the answer is 4! = 24 ways

Q12. How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 6, 7 if no digit is repeated?

Answer: For the number to be even, the units place must have an even digit.
The even digits available are 2, 4 and 6, so the units place has 3 choices.
After using one digit, 5 digits remain.
Tens place: 5 choices
Hundreds place: 4 choices
Total = 3 × 5 × 4 = 60 numbers

Related Study Material on Extramarks

• CBSE Important Questions Class 11 Maths Chapter 1
• CBSE Important Questions Class 11 Maths Chapter 2
• CBSE Important Questions Class 11 Maths Chapter 3
• CBSE Important Questions Class 11 Maths Chapter 4
• CBSE Important Questions Class 11 Maths Chapter 5
• CBSE Important Questions Class 11 Maths Chapter 7
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