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Class 9 · Advanced MathematicsNCERT 2026 Revised Syllabus

Chapter 5: Combinatorics

Class IX CBSE Advanced Mathematics

Introduces the fundamental principle of counting, factorials, permutations ($^{n}P_r$) and combinations ($^{n}C_r$), with word-arrangement and selection problems applied to everyday situations.

Find complete NCERT solutions for Class 9 Advanced Mathematics, Class IX CBSE Advanced Mathematics Chapter 5. Every question has a clear, step-by-step answer.

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Exercise 5.1(6 questions)

1

A restaurant offers 4 starters, 5 main courses and 3 desserts. How many different full meals (starter + main + dessert) are possible?

School Level Solution

Sol.
Solution:By the fundamental counting principle: $$4 \times 5 \times 3 = 60$$Answer: $\boxed{\text{60 ways}}$.
2

A hall has 5 doors to enter and 3 doors to exit. In how many ways can a person enter and exit?

School Level Solution

Sol.
Solution:$$5 \times 3 = 15$$Answer: $\boxed{\text{15 ways}}$.
3

A lock has 3 dials, each with digits 0-9, and repetition is allowed. How many combinations are possible?

School Level Solution

Sol.
Solution:$$10 \times 10 \times 10 = 1000$$Answer: $\boxed{\text{1000 combinations}}$.
4
Hard

How many numbers between 2000 and 3000 can be formed using the digits $\{2, 3, 4, 5, 6, 7\}$ without repetition?

School Level Solution

Sol.
Solution:Thousands digit must be 2 (1 choice). Remaining 3 places from the remaining 5 digits: $$5 \times 4 \times 3 = 60$$Total $= 1 \times 60 = 60$.Answer: $\boxed{\text{60 numbers}}$.
5
Hard

How many 3-digit even numbers can be formed from $\{1, 2, 3, 4, 6\}$? (a) without repetition, (b) with repetition allowed.

  1. a.Without repetition
  2. b.With repetition allowed

School Level Solution

Sol.
Solution:Units digit must be even: $2, 4, 6$ (3 choices).(a) Without repetition: Units (3 choices), Hundreds (4 remaining), Tens (3 remaining): $$3 \times 4 \times 3 = 36$$(b) With repetition: Units (3 choices), Hundreds (5 choices), Tens (5 choices): $$3 \times 5 \times 5 = 75$$Answer: $\boxed{\text{36 (no repetition); 75 (with repetition)}}$.
6
Hard

How many 3-digit numbers from 100 to 999 contain the digit 9 at (a) the unit\u2019s place, (b) the ten\u2019s place, (c) the hundred\u2019s place? Describe the pattern.

  1. a.9 at unit\u2019s place
  2. b.9 at ten\u2019s place
  3. c.9 at hundred\u2019s place

School Level Solution

Sol.
Solution:(a) 9 at unit\u2019s place: Hundreds (9 choices: 1-9), Tens (10 choices), Units fixed: $$9 \times 10 \times 1 = 90$$(b) 9 at ten\u2019s place: Hundreds (9 choices), Tens fixed, Units (10 choices): $$9 \times 1 \times 10 = 90$$(c) 9 at hundred\u2019s place: Hundreds fixed, Tens (10), Units (10): $$1 \times 10 \times 10 = 100$$Pattern: When 9 occupies the unit\u2019s or ten\u2019s place, the count is 90 each time (the hundred\u2019s place still excludes 0, giving 9 choices). But when 9 itself occupies the hundred\u2019s place, that restriction disappears and both remaining places get the full 10 choices, so the count rises to 100.Answer: $\boxed{\text{90, 90, 100 respectively}}$.

Exercise 5.2(4 questions)

1

Find the HCF and LCM of $6!$ and $5!$.

School Level Solution

Sol.
Solution:$$6! = 720, \quad 5! = 120$$Since $120 \mid 720$ (as $720 = 6 \times 120$): $$\text{HCF} = 120, \quad \text{LCM} = 720$$Answer: $\boxed{\text{HCF }= 120\text{, LCM }= 720}$.
2
Hard

Find $n$ if $\frac{n!}{(n-3)!} = 60$.

School Level Solution

Sol.
Solution:$$\frac{n!}{(n-3)!} = n(n-1)(n-2) = 60$$Trying $n = 5$: $5 \times 4 \times 3 = 60$ ✓.Answer: $\boxed{n = 5}$.
3
Hard

Find $x$ if $\frac{1}{5!} + \frac{1}{6!} = \frac{x}{7!}$.

School Level Solution

Sol.
Solution:Convert to sevenths of $7!$: $$\frac{1}{5!} = \frac{7 \times 6}{7!} = \frac{42}{7!}, \quad \frac{1}{6!} = \frac{7}{7!}$$$$x = \frac{7!}{5!} + \frac{7!}{6!} = (7 \times 6) + 7 = 42 + 7 = 49$$Answer: $\boxed{x = 49}$.
4
Hard

Solve: (a) $x! = 120$ (b) $(x!)^2 = 576$ (c) $(x!)^2 - 25(x!) + 24 = 0$ (d) $\frac{(x+2)! - (x+1)!}{x!} = 49$.

  1. a.$x! = 120$
  2. b.$(x!)^2 = 576$
  3. c.$(x!)^2 - 25(x!) + 24 = 0$
  4. d.$\frac{(x+2)! - (x+1)!}{x!} = 49$

School Level Solution

Sol.
Solution:(a) $x! = 120 \Rightarrow x = 5$.(b) $(x!)^2 = 576 \Rightarrow x! = 24 \Rightarrow x = 4$.(c) Let $t = x!$: $t^2 - 25t + 24 = 0 \Rightarrow (t-1)(t-24) = 0 \Rightarrow t = 1$ or $24$. $t = 1 \Rightarrow x = 0$ or $1$; $t = 24 \Rightarrow x = 4$. So $x = 0, 1, 4$.(d) $(x+2)! = (x+2)(x+1)x!$ and $(x+1)! = (x+1)x!$: $$\frac{(x+2)! - (x+1)!}{x!} = (x+1)[(x+2) - 1] = (x+1)^2 = 49 \Rightarrow x + 1 = \pm 7$$ Since $x \geq 0$, $x = 6$.Answer: $\boxed{\text{(a) }x = 5\text{, (b) }x = 4\text{, (c) }x = 0, 1, 4\text{, (d) }x = 6}$.

Exercise 5.3(6 questions)

1

In how many ways can 4 distinct cars be parked in 6 spaces?

School Level Solution

Sol.
Solution:$$^6P_4 = 6 \times 5 \times 4 \times 3 = 360$$Answer: $\boxed{\text{360 ways}}$.
2

How many 3-letter words can be formed from the letters of the word LOGIC?

School Level Solution

Sol.
Solution:LOGIC has 5 distinct letters: $$^5P_3 = 5 \times 4 \times 3 = 60$$Answer: $\boxed{\text{60 words}}$.
3
Hard

In how many ways can 3 maths books and 5 science books be arranged on a shelf so that all maths books come first?

School Level Solution

Sol.
Solution:Arrange 3 maths books in the first 3 positions: $3! = 6$. Arrange 5 science books in the remaining 5 positions: $5! = 120$.$$6 \times 120 = 720$$Answer: $\boxed{\text{720 ways}}$.
4
Hard

In how many ways can 5 boys and 2 girls be seated in a row so that the two girls always sit together?

School Level Solution

Sol.
Solution:Treat the 2 girls as one block: 6 units (5 boys + 1 block) arrange in $6! = 720$ ways. Arrange the girls within the block: $2! = 2$.$$720 \times 2 = 1440$$Answer: $\boxed{\text{1440 ways}}$.
5
Hard

How many 4-digit numbers greater than 6000 can be formed from $\{2, 4, 6, 8, 9\}$ without repetition?

School Level Solution

Sol.
Solution:Thousands digit must be $\{6, 8, 9\}$ (3 choices). Remaining 3 places from the remaining 4 digits: $$^4P_3 = 4 \times 3 \times 2 = 24$$$$3 \times 24 = 72$$Answer: $\boxed{\text{72 numbers}}$.
6
Hard

The word DAUGHTER has 8 letters (vowels A, U, E). (i) How many arrangements keep all the vowels together? (ii) How many arrangements keep the vowels not together?

  1. a.Vowels together
  2. b.Vowels not together

School Level Solution

Sol.
Solution:(i) Vowels together: treat A, U, E as one block → 6 units, arranged in $6! = 720$ ways. Vowels within block: $3! = 6$. $$720 \times 6 = 4320$$(ii) Total arrangements $= 8! = 40320$. $$\text{Not together} = 40320 - 4320 = 36000$$Answer: $\boxed{\text{(i) 4320 words, (ii) 36000 words}}$.

Exercise 5.4(6 questions)

1

In how many ways can a committee of 3 be chosen from 12 people?

School Level Solution

Sol.
Solution:$$^{12}C_3 = \frac{12!}{3! \, 9!} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220$$Answer: $\boxed{\text{220 ways}}$.
2
Hard

There are 12 points in a plane, of which 5 are collinear. How many triangles can be formed?

School Level Solution

Sol.
Solution:The 5 collinear points cannot form a triangle amongst themselves: $$\text{Triangles} = {}^{12}C_3 - {}^5C_3 = 220 - 10 = 210$$Answer: $\boxed{\text{210 triangles}}$.
3
Hard

A student must select 7 questions from a paper with Part A (7 questions) and Part B (5 questions), choosing at least 3 from each part. In how many ways?

School Level Solution

Sol.
Solution:Possible splits $(a, b)$ with $a + b = 7$, $a \geq 3$, $b \geq 3$: $(3, 4)$ or $(4, 3)$.$(3$ from A, $4$ from B): ${}^7C_3 \times {}^5C_4 = 35 \times 5 = 175$ $(4$ from A, $3$ from B): ${}^7C_4 \times {}^5C_3 = 35 \times 10 = 350$$$175 + 350 = 525$$Answer: $\boxed{\text{525 ways}}$.
4
Hard

Find the number of diagonals of a 10-sided polygon.

School Level Solution

Sol.
Solution:Number of diagonals of an $n$-sided polygon $= \frac{n(n-3)}{2}$.For $n = 10$: $$\frac{10 \times 7}{2} = 35$$Answer: $\boxed{\text{35 diagonals}}$.
5

If 15 people meet and each shakes hands with every other once, how many handshakes occur?

School Level Solution

Sol.
Solution:$$${}^{15}C_2 = \frac{15 \times 14}{2} = 105$$Answer: $\boxed{\text{105 handshakes}}$.
6
Hard

A committee of 3 is to be chosen from 2 men and 3 women. (a) How many committees are possible in total? (b) How many have exactly 1 man and 2 women?

  1. a.Total committees
  2. b.Exactly 1 man and 2 women

School Level Solution

Sol.
Solution:(a) Committee of 3 from $2 + 3 = 5$ people: $$${}^5C_3 = 10$$(b) Exactly 1 man and 2 women: $$${}^2C_1 \times {}^3C_2 = 2 \times 3 = 6$$Answer: $\boxed{\text{(a) 10 ways, (b) 6 ways}}$.

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