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NCERT Solutions · Class 11 Maths · Chapter 6

Chapter 6: Permutations and Combinations (Algebra)

Step-by-step solutions to all exercises of Chapter 6, Permutations and Combinations (NCERT Class 11 Maths, 2026-27): fundamental principle of counting, factorials, permutations with and without repetition, words from letters, and combinations. All 42 questions are answered, with the key answer highlighted.

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Multiplication principle: m ways then n ways gives mn ways. $^{n} P_{r} = \frac{n !}{( n - r )!}$ ; with repeated objects $\frac{n !}{p _{1} ! p _{2} ! \dots}$ ; $^{n} C_{r} = \frac{n !}{r ! ( n - r )!}$ , $^{n} C_{r} =^{n} C_{n - r}$ .

Exercise 6.1

How many 3-digit numbers can be formed from 1, 2, 3, 4, 5 if (i) repetition is allowed (ii) not allowed?

Solution

(i) 5 × 5 × 5 = 125. (ii) 5 × 4 × 3 = 60.

125; 60

How many 3-digit even numbers can be formed from 1, 2, 3, 4, 5, 6 if the digits can be repeated?

Solution

Units: 2, 4 or 6 (3 ways); tens and hundreds 6 ways each: 3 × 6 × 6 = 108.

108

How many 4-letter codes can be formed using the first 10 letters of the alphabet, no repetition?

Solution

10 × 9 × 8 × 7 = 5040.

5040

How many 5-digit telephone numbers can be made from 0–9 if each starts with 67 and no digit appears more than once?

Solution

67 _ _ _: the other three places from the remaining 8 digits: 8 × 7 × 6 = 336.

336

A coin is tossed 3 times and the outcomes recorded. How many possible outcomes?

Solution

2 × 2 × 2 = 8.

Given 5 flags of different colours, how many signals using 2 flags one below the other?

Solution

5 × 4 = 20.

Exercise 6.2

Evaluate (i) 8! (ii) 4! − 3!.

Solution

(i) 40320. (ii) 24 − 6 = 18.

40320; 18

Is 3! + 4! = 7!?

Solution

3! + 4! = 6 + 24 = 30, but 7! = 5040: No.

Compute 8!/(6! × 2!).

Solution

(8 × 7 × 6!)/(6! × 2) = 56/2 = 28.

If 1/6! + 1/7! = x/8!, find x.

Solution

Multiply by 8!: 8!/6! + 8!/7! = x ⇒ 56 + 8 = x = 64.

x = 64

Evaluate n!/(n − r)! for (i) n = 6, r = 2 (ii) n = 9, r = 5.

Solution

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

30; 15120

Exercise 6.3

How many 3-digit numbers from 1–9, no digit repeated?

Solution

⁹P₃ = 9 × 8 × 7 = 504.

504

How many 4-digit numbers have no digit repeated?

Solution

Thousands digit 1–9 (9 ways), then 9, 8, 7 (0 now allowed): 9 × 9 × 8 × 7 = 4536.

4536

How many 3-digit even numbers from 1, 2, 3, 4, 6, 7, no repetition?

Solution

Units: 2, 4 or 6 (3 ways); then 5 and 4: 3 × 5 × 4 = 60.

How many 4-digit numbers from 1, 2, 3, 4, 5, no repetition? How many are even?

Solution

⁵P₄ = 120. Even: units 2 or 4 (2 ways), the other three places ⁴P₃ = 24: 48.

120; 48

From a committee of 8, in how many ways can a chairman and a vice chairman be chosen?

Solution

⁸P₂ = 8 × 7 = 56.

Find n if ⁿ⁻¹P₃ : ⁿP₄ = 1 : 9.

Solution

(n − 1)(n − 2)(n − 3)/[n(n − 1)(n − 2)(n − 3)] = 1/n = 1/9 ⇒ n = 9.

n = 9

Find r if (i) ⁵Pᵣ = 2 ⁶Pᵣ₋₁ (ii) ⁵Pᵣ = ⁶Pᵣ₋₁.

Solution

⁵Pᵣ = 5!/(5 − r)!, ⁶Pᵣ₋₁ = 6!/(7 − r)!. (i) 5!/(5 − r)! = 2 × 6!/(7 − r)! ⇒ (7 − r)(6 − r) = 12 ⇒ r² − 13r + 30 = 0 ⇒ r = 3 or 10; r ≤ 5, so r = 3. (ii) (7 − r)(6 − r) = 6 ⇒ r² − 13r + 36 = 0 ⇒ r = 4 or 9: r = 4.

(i) r = 3; (ii) r = 4

How many words can be formed using all the letters of EQUATION, each exactly once?

Solution

8 different letters: 8! = 40320.

40320

How many words from MONDAY, no letter repeated, if (i) 4 letters are used (ii) all letters are used (iii) all letters are used but the first is a vowel?

Solution

(i) ⁶P₄ = 360. (ii) 6! = 720. (iii) first letter O or A (2 ways), the rest 5!: 240.

(i) 360; (ii) 720; (iii) 240

In how many of the distinct permutations of MISSISSIPPI do the four I's not come together?

Solution

M1, I4, S4, P2: total 11!/(4! 4! 2!) = 34650. Four I's together (one block): 8 objects with S4, P2: 8!/(4! 2!) = 840. Not together: 34650 − 840 = 33810.

33810

In how many ways can the letters of PERMUTATIONS be arranged if (i) words start with P and end with S (ii) vowels are all together (iii) there are always 4 letters between P and S?

Solution

12 letters, T twice. (i) Fix P and S: 10!/2! = 1814400. (ii) Vowels E, U, A, I, O as one block: 8 units (block + P, R, M, T, T, N, S): 8!/2! = 20160; inside 5! = 120: 2419200. (iii) P and S at positions (k, k + 5), k = 1, …, 7: 7 positions × 2 orders = 14; the other 10 letters: 10!/2! = 1814400: 14 × 1814400 = 25401600.

(i) 1814400; (ii) 2419200; (iii) 25401600

Exercise 6.4

If ⁿC₈ = ⁿC₂, find ⁿC₂.

Solution

8 ≠ 2, so n = 8 + 2 = 10; ¹⁰C₂ = 45.

Determine n if (i) ²ⁿC₃ : ⁿC₃ = 12 : 1 (ii) ²ⁿC₃ : ⁿC₃ = 11 : 1.

Solution

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

(i) n = 5; (ii) n = 6

How many chords can be drawn through 21 points on a circle?

Solution

Each chord joins 2 points: ²¹C₂ = 210.

210

In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?

Solution

⁵C₃ × ⁴C₃ = 10 × 4 = 40.

Find the number of ways of selecting 9 balls from 6 red, 5 white and 5 blue if each selection has 3 of each colour.

Solution

⁶C₃ × ⁵C₃ × ⁵C₃ = 20 × 10 × 10 = 2000.

2000

Number of 5-card combinations from 52 cards with exactly one ace.

Solution

One of the 4 aces and 4 of the 48 non-aces: ⁴C₁ × ⁴⁸C₄ = 4 × 194580 = 778320.

778320

A cricket team of 11 from 17 players, of whom 5 can bowl, must include exactly 4 bowlers. In how many ways?

Solution

⁵C₄ × ¹²C₇ = 5 × 792 = 3960.

3960

A bag has 5 black and 6 red balls. In how many ways can 2 black and 3 red balls be selected?

Solution

⁵C₂ × ⁶C₃ = 10 × 20 = 200.

200

A programme of 5 courses from 9, with 2 specific courses compulsory: in how many ways?

Solution

Choose the other 3 from the remaining 7: ⁷C₃ = 35.

Miscellaneous Exercise on Chapter 6

How many words, each of 2 vowels and 3 consonants, from DAUGHTER?

Solution

Vowels A, U, E; consonants D, G, H, T, R. ³C₂ × ⁵C₃ = 30 selections, each arranged in 5! = 120 ways: 3600.

3600

How many words using all the letters of EQUATION so that the vowels and consonants occur together?

Solution

Vowels E, U, A, I, O (block of 5) and consonants Q, T, N (block of 3): the two blocks in 2! ways, inside 5! and 3!: 2 × 120 × 6 = 1440.

1440

A committee of 7 from 9 boys and 4 girls with (i) exactly 3 girls (ii) at least 3 girls (iii) at most 3 girls.

Solution

(i) ⁴C₃ × ⁹C₄ = 4 × 126 = 504. (ii) 3 or 4 girls: 504 + ⁴C₄ × ⁹C₃ = 504 + 84 = 588. (iii) total ¹³C₇ = 1716 minus 4 girls (84): 1632.

(i) 504; (ii) 588; (iii) 1632

If the permutations of EXAMINATION are listed as in a dictionary, how many words come before the first word starting with E?

Solution

EXAMINATION: A2, I2, N2, E, X, M, T, O (11 letters). Only words beginning with A come before E: the other 10 letters (I2, N2): 10!/(2! 2!) = 907200.

907200

How many 6-digit numbers from 0, 1, 3, 5, 7, 9 are divisible by 10, no digit repeated?

Solution

Divisible by 10 ⇒ last digit 0; the other 5 digits fill 5 places: 5! = 120.

120

With 5 vowels and 21 consonants, how many words with two different vowels and two different consonants?

Solution

⁵C₂ × ²¹C₂ = 10 × 210 = 2100 selections, each arranged in 4! = 24 ways: 50400.

50400

A paper has 12 questions (Part I 5, Part II 7). Attempt 8, at least 3 from each part. In how many ways?

Solution

(3, 5): ⁵C₃⁷C₅ = 10 × 21 = 210; (4, 4): ⁵C₄⁷C₄ = 5 × 35 = 175; (5, 3): ⁵C₅⁷C₃ = 35. Total 420.

420

Number of 5-card combinations from 52 cards with exactly one king.

Solution

⁴C₁ × ⁴⁸C₄ = 778320.

778320

Seat 5 men and 4 women in a row so that the women occupy the even places.

Solution

Even places 2, 4, 6, 8 for the women: 4!; odd places for the men: 5!: 24 × 120 = 2880.

2880

From 25 students, 10 are chosen for a trip; 3 students will join only all together or none. In how many ways?

Solution

All 3 join: choose 7 more from 22: ²²C₇ = 170544; none of them: ²²C₁₀ = 646646. Total ²²C₇ + ²²C₁₀ = 817190.

817190

In how many ways can the letters of ASSASSINATION be arranged so that all the S's are together?

Solution

A3, S4, I2, N2, T, O (13 letters). SSSS as one block: 10 units with A3, I2, N2: 10!/(3! 2! 2!) = 151200.

151200

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