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Class 11 Maths NCERT Solutions

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Permutations and Combinations Class 11 Maths NCERT Solutions

The complete NCERT exercise solutions for Chapter 6, Permutations and Combinations — 31 questions from Ex 6.1 to Ex 6.4, each worked through step by step in the CBSE marking pattern. Arrangements, selections, circular permutations, arrangements with repetition, and the number of ways to divide objects.

Class:11Subject:MathsChapter:6
4 Key Formulas
DWritten byDeep Narayan
Updated
Key Concept Summary

How many questions are in NCERT Class 11 Maths Chapter 6?

Chapter 6 carries 4 exercise questions, numbered Ex 6.1 to Ex 6.4. All of them are solved step by step on this page, along with the chapter's key formulas and exam pointers.

01

Chapter Overview

This chapter is the counting toolbox of combinatorics. The fundamental principle of multiplication (and addition) lets you count compound events; factorials and the nPr / nCr formulas count arrangements and selections. The two big traps: telling a permutation (order matters) from a combination (order does not) and handling repetitions inside a word like MISSISSIPPI.

Board pattern

For word-arrangement questions, always divide by the factorial of each repeated letter. For selection questions, ask 'does the order matter?' — ordered → nPr, unordered → nCr. Write the formula out explicitly before evaluating so the examiner can award method marks.
02

Exercise 6.1 — Fundamental Principle of Counting

6Exercise questions

Step-by-step solution

  1. 1Each of the three places has 5 choices when repetition is allowed.
  2. 2(i) 5 × 5 × 5 = 125.
  3. 3(ii) First place 5 choices, second 4, third 3 → 5 × 4 × 3 = 60.

Final answer

(i) 125, (ii) 60.

Step-by-step solution

  1. 1The unit's digit must be even: 2, 4 or 6 → 3 choices.
  2. 2Hundred's and ten's places: 6 choices each (repetition allowed).
  3. 3Total = 6 × 6 × 3 = 108.

Final answer

108.

Step-by-step solution

  1. 1First letter: 10 choices, then 9, 8, 7 for the next places.
  2. 2Total = 10 × 9 × 8 × 7 = 5040.

Final answer

5040.

Step-by-step solution

  1. 1The first two digits are fixed as 6 and 7.
  2. 2Remaining digits available: 8 (0, 1, 2, 3, 4, 5, 8, 9).
  3. 3Last three places: 8 × 7 × 6 = 336.

Final answer

336.

Step-by-step solution

  1. 1Each toss has 2 outcomes (Head or Tail).
  2. 2Total = 2 × 2 × 2 = 8.

Final answer

8.

Step-by-step solution

  1. 1Top flag: 5 choices, bottom flag: 4 choices (order matters).
  2. 2Total = 5 × 4 = 20.

Final answer

20.

03

Exercise 6.2 — Factorials and n! / (n − r)!

5Exercise questions

Step-by-step solution

  1. 1(i) 8! = 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 = 40320.
  2. 2(ii) 4! − 3! = 24 − 6 = 18.

Final answer

(i) 40320, (ii) 18.

Step-by-step solution

  1. 13! + 4! = 6 + 24 = 30.
  2. 27! = 5040.
  3. 330 ≠ 5040, so the statement is false.

Final answer

No (30 ≠ 5040).

Step-by-step solution

  1. 18! / (6! × 2!) = (8 × 7 × 6!) / (6! × 2).
  2. 2= 8 × 7 / 2 = 28.

Final answer

28.

Step-by-step solution

  1. 11/6! + 1/7! = 1/6! + 1/(7 × 6!) = (1 + 1/7)/6! = (8/7)/6! = 8/7!.
  2. 2Rewrite right side: x/8! = x/(8 × 7!).
  3. 3(8 × 7!)/8! × 8/7! → x = 8 × 8 = 64.

Final answer

x = 64.

Step-by-step solution

  1. 1(i) 6!/4! = 6 × 5 × 4!/4! = 6 × 5 = 30.
  2. 2(ii) 9!/4! = 9 × 8 × 7 × 6 × 5 × 4!/4! = 15120.

Final answer

(i) 30, (ii) 15120.

04

Exercise 6.3 — Permutations

11Exercise questions

Step-by-step solution

  1. 1This is the number of permutations of 9 digits taken 3 at a time: ⁹P₃.
  2. 2= 9 × 8 × 7 = 504.

Final answer

504.

Step-by-step solution

  1. 1Thousands place: 9 choices (1 to 9, since 0 cannot lead).
  2. 2Hundreds place: 9 choices (0 plus 8 remaining digits).
  3. 3Tens: 8 choices, Units: 7 choices.
  4. 4Total = 9 × 9 × 8 × 7 = 4536.

Final answer

4536.

Step-by-step solution

  1. 1For an even number, unit's digit is 2, 4 or 6 → 3 choices.
  2. 2Hundred's digit: 5 choices (remaining digits).
  3. 3Ten's digit: 4 choices.
  4. 4Total = 3 × 5 × 4 = 60.

Final answer

60.

Step-by-step solution

  1. 1Total 4-digit numbers: ⁵P₄ = 5 × 4 × 3 × 2 = 120.
  2. 2Even numbers: unit's digit is 2 or 4 → 2 choices.
  3. 3Remaining 3 places from 4 digits: 4 × 3 × 2 = 24 ways each case.
  4. 4Even numbers = 2 × 24 = 48.

Final answer

120 total, 48 even.

Step-by-step solution

  1. 1Chairman: 8 choices.
  2. 2Vice chairman: 7 choices (excluding the chairman).
  3. 3Total = 8 × 7 = 56.

Final answer

56.

Step-by-step solution

  1. 1ⁿ⁻¹P₃ = (n−1)(n−2)(n−3), ⁿP₄ = n(n−1)(n−2)(n−3).
  2. 2Ratio = 1/n = 1/9.
  3. 3n = 9.

Final answer

n = 9.

Step-by-step solution

  1. 1(i) 5!/(5−r)! = 2 · 6!/(7−r)! → (7−r)(6−r) = 12.
  2. 2r² − 13r + 30 = 0 → (r − 3)(r − 10) = 0 → r = 3 (r = 10 is impossible).
  3. 3(ii) 5!/(5−r)! = 6!/(7−r)! → (7−r)(6−r) = 6.
  4. 4r² − 13r + 36 = 0 → (r − 4)(r − 9) = 0 → r = 4.

Final answer

(i) r = 3, (ii) r = 4.

Step-by-step solution

  1. 1EQUATION has 8 distinct letters.
  2. 2Number of arrangements = 8! = 40320.

Final answer

40320.

Step-by-step solution

  1. 1MONDAY has 6 distinct letters.
  2. 2(i) ⁶P₄ = 6 × 5 × 4 × 3 = 360.
  3. 3(ii) 6! = 720.
  4. 4(iii) First letter is O or A → 2 choices, remaining 5 letters in 5! = 120 ways.
  5. 5Total = 2 × 120 = 240.

Final answer

(i) 360, (ii) 720, (iii) 240.

Step-by-step solution

  1. 1MISSISSIPPI has 11 letters: I×4, S×4, P×2, M×1.
  2. 2Total distinct arrangements = 11!/(4!·4!·2!) = 34650.
  3. 3All four I's together: treat IIII as one block → 8 objects: 8!/(4!·2!) = 840.
  4. 4Not together = 34650 − 840 = 33810.

Final answer

33810.

Step-by-step solution

  1. 1PERMUTATIONS has 12 letters with T repeated twice.
  2. 2(i) Fix P first, S last; middle 10 letters with T twice: 10!/2! = 1814400.
  3. 3(ii) Vowels E, U, A, I, O form one block g with 7 consonants → 8 objects, T twice.
  4. 48!/2! × 5! = 20160 × 120 = 2419200.
  5. 5(iii) Pairs of positions for P, S with exactly 4 letters between: (1,6), (2,7), …, (7,12) → 7 pairs, each swappable.
  6. 67 × 2 × 10!/2! = 14 × 1814400 = 25401600.

Final answer

(i) 1814400, (ii) 2419200, (iii) 25401600.

05

Exercise 6.4 — Combinations

9Exercise questions

Step-by-step solution

  1. 1ⁿC₈ = ⁿC₂ ⇒ n = 8 + 2 = 10.
  2. 2¹⁰C₂ = 10 × 9 / 2 = 45.

Final answer

45.

Step-by-step solution

  1. 1²ⁿC₃/ⁿC₃ = [2n(2n−1)(2n−2)/6] / [n(n−1)(n−2)/6] = 4(2n−1)/(n−2).
  2. 2(i) 4(2n−1)/(n−2) = 12 → 8n − 4 = 12n − 24 → n = 5.
  3. 3(ii) 4(2n−1)/(n−2) = 11 → 8n − 4 = 11n − 22 → n = 6.

Final answer

(i) n = 5, (ii) n = 6.

Step-by-step solution

  1. 1A chord joins 2 points (order does not matter).
  2. 2Number = ²¹C₂ = 21 × 20 / 2 = 210.

Final answer

210.

Step-by-step solution

  1. 1Choose 3 boys from 5: ⁵C₃ = 10.
  2. 2Choose 3 girls from 4: ⁴C₃ = 4.
  3. 3Total = 10 × 4 = 40.

Final answer

40.

Step-by-step solution

  1. 1Select 3 red from 6: ⁶C₃ = 20.
  2. 2Select 3 white from 5: ⁵C₃ = 10.
  3. 3Select 3 blue from 5: ⁵C₃ = 10.
  4. 4Total = 20 × 10 × 10 = 2000.

Final answer

2000.

Step-by-step solution

  1. 1Choose 1 ace from 4: ⁴C₁ = 4.
  2. 2Choose remaining 4 cards from the 48 non-aces: ⁴⁸C₄ = 194580.
  3. 3Total = 4 × 194580 = 778320.

Final answer

778320.

Step-by-step solution

  1. 1Choose 4 bowlers from 5: ⁵C₄ = 5.
  2. 2Choose remaining 7 players from the other 12: ¹²C₇ = 792.
  3. 3Total = 5 × 792 = 3960.

Final answer

3960.

Step-by-step solution

  1. 1Choose 2 black from 5: ⁵C₂ = 10.
  2. 2Choose 3 red from 6: ⁶C₃ = 20.
  3. 3Total = 10 × 20 = 200.

Final answer

200.

Step-by-step solution

  1. 1The 2 compulsory courses are fixed.
  2. 2Choose the remaining 3 courses from the other 7: ⁷C₃ = 35.

Final answer

35.

Quick Revision

Key formulas at a glance

Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.

Permutation

Combination

Arrangements with repetition

Circular permutation

Exam Strategy

How this chapter is asked

High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.

  • Order matters in a permutation and not in a combination — if swapping two chosen items changes the outcome, it is a permutation.
  • In circular arrangements fix one object first to remove the rotational duplication, which is why the count is (n−1)! rather than n!.

FAQ

Frequently asked questions

How many questions are in NCERT Class 11 Maths Chapter 6 (Permutations and Combinations)?

There are 4 exercise questions in this chapter, numbered Ex 6.1 to Ex 6.4. Every one is solved step by step on this page in the official NCERT numbering.

Which formulas come up in Permutations and Combinations Class 11 Maths?

The formulas this chapter's questions actually turn on are: Permutation, Combination, Arrangements with repetition, Circular permutation. They are listed with their expressions in the key formulas section below, and the solved questions show where each one is used.

Is Permutations and Combinations important for JEE Main?

Very important — permutations and combinations are a guaranteed unit in boards and JEE Main, and the counting-principle questions are high-frequency.

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