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.

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TL;DR: Free step-by-step NCERT solutions for Class 11 Mathematics chapter "Permutations and Combinations" — 8 important questions with detailed answers for C…

Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated Jul 23, 2026

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Key Questions Covered:

Find the value of 8P3 and 8C3.
In how many ways can 5 people be arranged in a row?
A committee of 4 people is to be selected from a group of 10 people. In how m…
In how many ways can the letters of the word MATHEMATICS be arranged?
In how many ways can 3 boys and 2 girls be arranged in a row such that the gi…
In how many ways can 4 red balls, 3 blue balls, and 2 green balls be arranged…
+ 2 more questions in the full chapter

Solutions Summary:

Question	Status
Find the value of 8P3 and 8C3.	✓ Solved
In how many ways can 5 people be arranged in a row?	✓ Solved
A committee of 4 people is to be selected from a group of…	✓ Solved
In how many ways can the letters of the word MATHEMATICS …	✓ Solved
In how many ways can 3 boys and 2 girls be arranged in a …	✓ Solved
In how many ways can 4 red balls, 3 blue balls, and 2 gre…	✓ Solved

Showing 6 of 8 questions

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,...

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) Positio...

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 b...

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!...

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,...

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 numer...

Showing 6 of 8 questions. Visit the full page for complete solutions.