Permutations and Combinations — Class 11 Mathematics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 11 Mathematics chapter "Permutations and Combinations" — 8 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 11 Mathematics chapter "Permutations and Combinations" — 8 important questions with detailed answers for C…
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NCERT Solutions for Class 11 Mathematics — Permutations and Combinations. Step-by-step answers to all 8 textbook questions from this chapter, written for CBSE board preparation and free to use.
Permutations and Combinations — All 8 Questions Solved
Q1. Find the value of 8P3 and 8C3.
Step 1: Recall the permutation formula.
nPr = n! / (n - r)!
Step 2: Calculate 8P3.
8P3 = 8! / (8 - 3)!
= 8! / 5!
= (8 × 7 × 6 × 5!) / 5!
= 8 × 7 × 6
= 56 × 6
= 336
Step 3: Recall the combination formula.
nCr = n! / [r!(n - r)!]
Step 4: Calculate 8C3.
8C3 = 8! / [3!(8 - 3)!]
= 8! / (3! × 5!)
= (8 × 7 × 6 × 5!) / (3! × 5!)
= (8 × 7 × 6) / (3 × 2 × 1)
= 336 / 6
= 56
Step 5: Relationship between permutations and combinations.
nPr = nCr × r!
8P3 = 8C3 × 3!
336 = 56 × 6 ✓
Final Answer: 8P3 = 336, 8C3 = 56
Q2. In how many ways can 5 people be arranged in a row?
Step 1: Understand the problem.
We need to arrange 5 distinct people in 5 positions.
This is a permutation problem where order matters.
Step 2: Apply the permutation formula.
nPn = n! (all n objects arranged in all n positions)
Step 3: Calculate 5P5 = 5!
5! = 5 × 4 × 3 × 2 × 1
Step 4: Compute step by step.
5 × 4 = 20
20 × 3 = 60
60 × 2 = 120
120 × 1 = 120
Step 5: Verify by reasoning.
Position 1: 5 choices
Position 2: 4 choices (one person used)
Position 3: 3 choices (two people used)
Position 4: 2 choices (three people used)
Position 5: 1 choice (four people used)
Total = 5 × 4 × 3 × 2 × 1 = 120
Final Answer: 5P5 = 5! = 120 ways
Q3. A committee of 4 people is to be selected from a group of 10 people. In how many ways can this be done?
Step 1: Understand the problem.
We need to select 4 people from 10.
Order does not matter (a committee {A, B, C, D} is the same as {B, A, D, C}).
This is a combination problem.
Step 2: Apply the combination formula.
10C4 = 10! / [4!(10 - 4)!]
= 10! / (4! × 6!)
Step 3: Simplify.
10C4 = (10 × 9 × 8 × 7 × 6!) / (4! × 6!)
= (10 × 9 × 8 × 7) / (4!)
= (10 × 9 × 8 × 7) / (4 × 3 × 2 × 1)
= (10 × 9 × 8 × 7) / 24
Step 4: Calculate the numerator.
10 × 9 = 90
90 × 8 = 720
720 × 7 = 5040
Step 5: Divide by denominator.
5040 / 24 = 210
Step 6: Verify using the formula for combinations.
10C4 = (10 × 9 × 8 × 7) / (4 × 3 × 2 × 1)
Cross-cancel: (10 × 9 × 8 × 7) / (4 × 3 × 2 × 1)
= (5 × 3 × 4 × 7) / 1 = 420 / 2 = 210 ✓
Final Answer: 10C4 = 210 ways
Q4. In how many ways can the letters of the word MATHEMATICS be arranged?
Step 1: Count the letters in MATHEMATICS.
M-A-T-H-E-M-A-T-I-C-S
Total letters = 11
Step 2: Count the frequency of each letter.
M: 2
A: 2
T: 2
H: 1
E: 1
I: 1
C: 1
S: 1
Step 3: Apply the formula for permutations with repetition.
When some objects are identical, the number of arrangements is:
n! / (n₁! × n₂! × ... × nₖ!)
where n is the total number of objects and n₁, n₂, ..., nₖ are the frequencies of identical objects.
Step 4: Calculate.
11! / (2! × 2! × 2! × 1! × 1! × 1! × 1! × 1!)
= 11! / (2! × 2! × 2!)
= 11! / (2 × 2 × 2)
= 11! / 8
Step 5: Compute 11!
11! = 39,916,800
Step 6: Divide by 8.
39,916,800 / 8 = 4,989,600
Final Answer: The letters of MATHEMATICS can be arranged in 4,989,600 ways.
Q5. In how many ways can 3 boys and 2 girls be arranged in a row such that the girls are together?
Step 1: Treat the 2 girls as a single unit.
We now have 3 boys + 1 unit of girls = 4 units to arrange.
Step 2: Arrange these 4 units in a row.
4P4 = 4! = 24 ways
Step 3: Arrange the 2 girls within their unit.
The 2 girls can be arranged among themselves in:
2P2 = 2! = 2 ways
Step 4: Apply the multiplication principle.
Total arrangements = (Arrangements of 4 units) × (Arrangements of girls within their unit)
= 24 × 2
= 48
Step 5: Verify with an example.
Let boys be B₁, B₂, B₃ and girls be G₁, G₂.
One arrangement: (G₁G₂)B₁B₂B₃
Another: (G₂G₁)B₁B₂B₃
These are counted as different, which is correct.
Final Answer: 48 ways
Q6. In how many ways can 4 red balls, 3 blue balls, and 2 green balls be arranged in a row?
Step 1: Count total balls.
4 + 3 + 2 = 9 balls
Step 2: Count frequency of each color.
Red: 4 (identical)
Blue: 3 (identical)
Green: 2 (identical)
Step 3: Apply the permutation formula for objects with repetition.
Number of arrangements = n! / (n₁! × n₂! × n₃!)
where n = 9, n₁ = 4, n₂ = 3, n₃ = 2
Step 4: Calculate.
9! / (4! × 3! × 2!)
= (9 × 8 × 7 × 6 × 5 × 4!) / (4! × 3! × 2!)
= (9 × 8 × 7 × 6 × 5) / (3! × 2!)
= (9 × 8 × 7 × 6 × 5) / (6 × 2)
= (9 × 8 × 7 × 6 × 5) / 12
Step 5: Calculate numerator.
9 × 8 = 72
72 × 7 = 504
504 × 6 = 3024
3024 × 5 = 15,120
Step 6: Divide by 12.
15,120 / 12 = 1260
Final Answer: The balls can be arranged in 1260 ways.
Q7. From 6 men and 4 women, in how many ways can a committee of 5 be selected such that there are at least 2 women?
Step 1: Identify the constraint.
We need a committee of 5 people with at least 2 women.
This means: 2 women + 3 men, OR 3 women + 2 men, OR 4 women + 1 man
Step 2: Case 1 - 2 women and 3 men.
Ways = 4C2 × 6C3
= [4! / (2! × 2!)] × [6! / (3! × 3!)]
= [4 × 3 / 2] × [6 × 5 × 4 / (3 × 2 × 1)]
= 6 × 20
= 120
Step 3: Case 2 - 3 women and 2 men.
Ways = 4C3 × 6C2
= [4! / (3! × 1!)] × [6! / (2! × 4!)]
= 4 × [6 × 5 / 2]
= 4 × 15
= 60
Step 4: Case 3 - 4 women and 1 man.
Ways = 4C4 × 6C1
= 1 × 6
= 6
Step 5: Apply the addition principle.
Total ways = Case 1 + Case 2 + Case 3
= 120 + 60 + 6
= 186
Step 6: Verify using complementary counting (optional).
Total committees with no restriction: 10C5 = 252
Committees with 0 or 1 woman:
- 0 women, 5 men: 4C0 × 6C5 = 1 × 6 = 6
- 1 woman, 4 men: 4C1 × 6C4 = 4 × 15 = 60
Total with < 2 women = 6 + 60 = 66
Committees with ≥ 2 women = 252 - 66 = 186 ✓
Final Answer: 186 ways
Q8. How many numbers between 100 and 1000 can be formed using the digits 0, 1, 2, 3, 4, 5 without repetition?
Step 1: Understand the requirement.
We need 3-digit numbers between 100 and 1000.
Format: _ _ _ (hundreds, tens, units)
Step 2: Apply the constraint 'without repetition'.
Each digit can be used at most once.
The first digit cannot be 0 (otherwise it won't be a 3-digit number).
Step 3: Count choices for the hundreds place.
Available digits: 1, 2, 3, 4, 5 (not 0)
Choices = 5
Step 4: Count choices for the tens place.
After using one digit for hundreds, we have 5 remaining digits (including 0 now).
Choices = 5
Step 5: Count choices for the units place.
After using two digits, we have 4 remaining digits.
Choices = 4
Step 6: Apply the multiplication principle.
Total numbers = 5 × 5 × 4 = 100
Step 7: Verify with reasoning.
Hundreds: {1, 2, 3, 4, 5} → 5 ways
Tens: {0, 1, 2, 3, 4, 5} - (one digit used) → 5 ways
Units: {0, 1, 2, 3, 4, 5} - (two digits used) → 4 ways
Total = 5 × 5 × 4 = 100 ✓
Final Answer: 100 numbers can be formed.
Permutations and Combinations — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. How many 3-digit numbers can be formed using the digits 1, 2, 3, 4, 5 if repetition of digits is allowed?
- A. 60
- B. 120
- C. 125
- D. 243
Answer: C. 125 — For each of the three positions (hundreds, tens, units), there are 5 choices, and repetition is allowed. So, by the Fundamental Principle of Counting, total numbers = 5 × 5 × 5 = 125.
MCQ 2. What is the value of 8! / 6!?
- A. 1/56
- B. 56
- C. 8
- D. 48
Answer: B. 56 — 8! / 6! = (8 × 7 × 6!) / 6! = 8 × 7 = 56.
MCQ 3. Calculate the value of P(7, 2).
- A. 14
- B. 21
- C. 42
- D. 49
Answer: C. 42 — The permutation P(n, r) is given by n! / (n-r)!. So, P(7, 2) = 7! / (7-2)! = 7! / 5! = 7 × 6 = 42.
MCQ 4. Calculate the value of C(10, 3).
- A. 10
- B. 30
- C. 120
- D. 720
Answer: C. 120 — The combination C(n, r) is given by n! / (r! × (n-r)!). So, C(10, 3) = 10! / (3! × 7!) = (10 × 9 × 8) / (3 × 2 × 1) = 120.
MCQ 5. A coin is tossed 4 times. How many possible outcomes are there?
- A. 4
- B. 8
- C. 16
- D. 32
Answer: C. 16 — Each toss has 2 possible outcomes (Heads or Tails). Since there are 4 independent tosses, the total number of outcomes is 2 × 2 × 2 × 2 = 16.
MCQ 6. How many distinct permutations can be made from the letters of the word "MISSISSIPPI"?
- A. 34650
- B. 1260
- C. 3465
- D. 6930
Answer: A. 34650 — The word MISSISSIPPI has 11 letters: M (1), I (4), S (4), P (2). The number of distinct permutations is 11! / (4! × 4! × 2!) = 34650.
MCQ 7. How many 4-digit numbers can be formed using the digits 0, 1, 2, 3, 4 if repetition of digits is not allowed?
- A. 120
- B. 96
- C. 24
- D. 100
Answer: B. 96 — The first digit cannot be 0 (4 choices: 1, 2, 3, 4). The second digit can be any of the remaining 4 digits (including 0). The third has 3 choices, and the fourth has 2 choices. So, 4 × 4 × 3 × 2 = 96.
MCQ 8. If C(n, 8) = C(n, 2), then what is the value of C(n, 2)?
- A. 10
- B. 45
- C. 90
- D. 55
Answer: B. 45 — If C(n, x) = C(n, y), then x = y or x + y = n. Since 8 ≠ 2, we have n = 8 + 2 = 10. Thus, C(n, 2) = C(10, 2) = (10 × 9) / (2 × 1) = 45.
MCQ 9. In how many ways can a committee of 3 men and 2 women be formed from a group of 7 men and 5 women?
- A. 350
- B. 35
- C. 10
- D. 120
Answer: A. 350 — Number of ways to select 3 men from 7 is C(7, 3) = 35. Number of ways to select 2 women from 5 is C(5, 2) = 10. Total ways = C(7, 3) × C(5, 2) = 35 × 10 = 350.
MCQ 10. How many straight lines can be formed by joining 10 non-collinear points?
- A. 10
- B. 20
- C. 45
- D. 90
Answer: C. 45 — A straight line is formed by choosing any 2 points from the 10. Since the points are non-collinear, any pair forms a unique line. So, C(10, 2) = (10 × 9) / (2 × 1) = 45.
MCQ 11. If P(n, 4) = 12 × P(n, 2), find the value of n.
- A. 6
- B. 5
- C. 7
- D. 8
Answer: A. 6 — P(n, 4) = n(n-1)(n-2)(n-3) and P(n, 2) = n(n-1). So, n(n-1)(n-2)(n-3) = 12 × n(n-1). Dividing by n(n-1) (since n≥4), we get (n-2)(n-3) = 12, which simplifies to n² - 5n - 6 = 0. Factoring gives (n-6)(n+1) = 0. Since n must be positive, n = 6.
MCQ 12. In how many ways can the letters of the word "DAUGHTER" be arranged so that the vowels always come together?
- A. 720
- B. 1440
- C. 4320
- D. 5040
Answer: C. 4320 — The word DAUGHTER has 8 distinct letters. Vowels are A, U, E (3 vowels). Treat these 3 vowels as a single block. Now we arrange 5 consonants + 1 vowel block = 6 units in 6! ways. The 3 vowels within their block can be arranged in 3! ways. Total arrangements = 6! × 3! = 720 × 6 = 4320.
MCQ 13. A group consists of 4 boys and 7 girls. In how many ways can a team of 4 members be selected if the team has at least 3 girls?
- A. 70
- B. 105
- C. 140
- D. 175
Answer: D. 175 — The team must have at least 3 girls. This means we can have either 3 girls and 1 boy (C(7,3) × C(4,1) = 35 × 4 = 140 ways) or 4 girls and 0 boys (C(7,4) × C(4,0) = 35 × 1 = 35 ways). The total number of ways is 140 + 35 = 175.
MCQ 14. How many 5-digit numbers can be formed using the digits 1, 2, 3, 4, 5, 6, 7, 8, 9 (without repetition) such that the number is divisible by 5?
- A. 3360
- B. 1680
- C. 840
- D. 420
Answer: B. 1680 — For a number to be divisible by 5, its last digit must be 5 (since 0 is not available). So, the last digit is fixed (1 way). The remaining 4 digits must be chosen from the remaining 8 digits (1, 2, 3, 4, 6, 7, 8, 9) and arranged in the first 4 positions. This is P(8, 4) = 8 × 7 × 6 × 5 = 1680. Total ways = 1 × 1680 = 1680.
MCQ 15. There are 12 points in a plane, out of which 5 are collinear. How many triangles can be formed using these points?
- A. 185
- B. 210
- C. 220
- D. 250
Answer: B. 210 — Total number of triangles from 12 points without collinearity is C(12, 3) = 220. From the 5 collinear points, C(5, 3) = 10 combinations would not form triangles. So, total triangles = 220 - 10 = 210.
MCQ 16. A question paper has two parts, Part A and Part B. Part A has 6 questions (True/False) and Part B has 5 questions (MCQ A, B, C). How many ways can a student answer all questions?
- A. 11
- B. 30
- C. 64 * 243
- D. 64 + 243
Answer: C. 64 * 243 — For Part A, there are 2 choices for each of the 6 questions, so 2^6 ways. For Part B, there are 3 choices for each of the 5 questions, so 3^5 ways. Total ways = 2^6 × 3^5 = 64 × 243.
MCQ 17. What is 0! equal to?
- A. 0
- B. 1
- C. Undefined
- D. Infinity
Answer: B. 1 — By mathematical definition and convention, 0! (zero factorial) is equal to 1. This is essential for the consistency of permutation and combination formulas.
MCQ 18. If C(n, r) is the number of combinations, which of the following is true?
- A. C(n, r) = C(n, n-r)
- B. C(n, r) = C(r, n)
- C. C(n, r) = n! / r!
- D. C(n, r) = nC_r + nC_{r+1}
Answer: A. C(n, r) = C(n, n-r) — A fundamental property of combinations is C(n, r) = C(n, n-r), meaning choosing r items is equivalent to choosing (n-r) items to leave behind.
MCQ 19. How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6, if repetition of digits is allowed?
- A. 108
- B. 60
- C. 18
- D. 216
Answer: A. 108 — For an even number, the unit digit must be even (2, 4, 6 - 3 choices). For the hundreds and tens digits, any of the 6 digits can be used as repetition is allowed. So, 6 × 6 × 3 = 108.
MCQ 20. In how many ways can 5 boys and 4 girls be arranged in a row so that all the girls sit together?
- A. 5! * 4!
- B. 6! * 4!
- C. 9!
- D. 5! * 5!
Answer: B. 6! * 4! — Treat the 4 girls as a single block. Now we arrange 5 boys + 1 block of girls = 6 units in 6! ways. The 4 girls within their block can be arranged in 4! ways. Total arrangements = 6! × 4!.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.