Last Updated on November 24, 2025 by Aman Singh
This exercise focuses on combinations, which is the selection of objects where the order does not matter. The formula for the number of combinations of $n$ distinct objects taken $r$ at a time is $C(n, r) = \frac{n!}{r!(n-r)!}$.
1. If $C(n, 8) = C(n, 2)$, find $C(n, 2)$.
We use the property that if $C(n, a) = C(n, b)$, then either $a = b$ or $a + b = n$.
Since $8 \ne 2$, we must have:
$$n = 8 + 2$$
$$n = 10$$
Now find $C(n, 2)$, which is $C(10, 2)$:
$$C(10, 2) = \frac{10!}{2!(10-2)!} = \frac{10!}{2!8!}$$
$$C(10, 2) = \frac{10 \times 9 \times 8!}{2 \times 1 \times 8!} = \frac{10 \times 9}{2} = 5 \times 9 = \mathbf{45}$$
2. Determine $n$
(i) $C(2n, 3) : C(n, 3) = 12 : 1$
$$\frac{C(2n, 3)}{C(n, 3)} = \frac{12}{1}$$
Expand the combination terms:
$$\frac{\frac{2n(2n-1)(2n-2)}{3 \times 2 \times 1}}{\frac{n(n-1)(n-2)}{3 \times 2 \times 1}} = 12$$
$$\frac{2n(2n-1)(2n-2)}{n(n-1)(n-2)} = 12$$
Factor out common terms. Note that $2n-2 = 2(n-1)$:
$$\frac{2n(2n-1)2(n-1)}{n(n-1)(n-2)} = 12$$
Cancel $n$ (assuming $n \ne 0$) and $(n-1)$ (assuming $n \ne 1$):
$$\frac{4(2n-1)}{n-2} = 12$$
Divide both sides by 4:
$$\frac{2n-1}{n-2} = 3$$
$$2n – 1 = 3(n – 2)$$
$$2n – 1 = 3n – 6$$
$$6 – 1 = 3n – 2n$$
$$n = \mathbf{5}$$
(ii) $C(2n, 3) : C(n, 3) = 11 : 1$
$$\frac{C(2n, 3)}{C(n, 3)} = 11$$
Using the simplification from part (i):
$$\frac{4(2n-1)}{n-2} = 11$$
$$4(2n – 1) = 11(n – 2)$$
$$8n – 4 = 11n – 22$$
$$22 – 4 = 11n – 8n$$
$$18 = 3n$$
$$n = \frac{18}{3}$$
$$n = \mathbf{6}$$
3. How many chords can be drawn through 21 points on a circle?
A chord is formed by joining any two distinct points on the circle. Since the order of the points does not matter (point A joined to B is the same chord as B joined to A), this is a combination problem.
We choose $r=2$ points from $n=21$ points.
$$\text{Number of chords} = C(21, 2)$$
$$C(21, 2) = \frac{21 \times 20}{2 \times 1} = 21 \times 10 = \mathbf{210}$$
4. Selecting a team of 3 boys and 3 girls from 5 boys and 4 girls.
The selections of boys and girls are independent events. We use the multiplication principle.
- Select 3 boys from 5 boys:$$C(5, 3) = \frac{5!}{3!2!} = \frac{5 \times 4}{2 \times 1} = 10 \text{ ways}$$
- Select 3 girls from 4 girls:$$C(4, 3) = \frac{4!}{3!1!} = \frac{4}{1} = 4 \text{ ways}$$
$$\text{Total ways} = C(5, 3) \times C(4, 3)$$
$$\text{Total ways} = 10 \times 4 = \mathbf{40}$$
5. Selecting 9 balls (3 of each colour) from 6 red, 5 white, and 5 blue balls.
We need to make three independent selections:
- Select 3 red balls from 6 red balls:$$C(6, 3) = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \text{ ways}$$
- Select 3 white balls from 5 white balls:$$C(5, 3) = \frac{5 \times 4}{2 \times 1} = 10 \text{ ways}$$
- Select 3 blue balls from 5 blue balls:$$C(5, 3) = 10 \text{ ways}$$
$$\text{Total ways} = C(6, 3) \times C(5, 3) \times C(5, 3)$$
$$\text{Total ways} = 20 \times 10 \times 10 = \mathbf{2000}$$
6. 5-card combinations with exactly one ace.
A standard deck has 52 cards, including 4 Aces. The remaining non-Ace cards are $52 – 4 = 48$. We must select 5 cards in total.
- Select exactly 1 Ace from 4 Aces:$$C(4, 1) = 4 \text{ ways}$$
- Select 4 non-Ace cards from 48 non-Ace cards:$$C(48, 4) = \frac{48 \times 47 \times 46 \times 45}{4 \times 3 \times 2 \times 1}$$$$C(48, 4) = 4 \times 47 \times 46 \times 15 = 194,580 \text{ ways}$$
$$\text{Total combinations} = C(4, 1) \times C(48, 4)$$
$$\text{Total combinations} = 4 \times 194,580 = \mathbf{778,320}$$
7. Selecting a cricket team of 11 with exactly 4 bowlers.
Total players: 17. Bowlers: 5. Non-bowlers (Batsmen/Wicket-keepers): $17 – 5 = 12$.
The team must have 11 players, including exactly 4 bowlers.
- Select exactly 4 bowlers from 5 available bowlers:$$C(5, 4) = C(5, 1) = 5 \text{ ways}$$
- Select the remaining 7 players ($11 – 4$) from the 12 non-bowlers:$$C(12, 7) = \frac{12!}{7!5!} = \frac{12 \times 11 \times 10 \times 9 \times 8}{5 \times 4 \times 3 \times 2 \times 1}$$$$C(12, 7) = 12 \times 11 \times 2 \times 9 \times 8 / (120) = 12 \times 11 \times 2 \times 9 \times 8 / 120 = 792 \text{ ways}$$
$$\text{Total ways} = C(5, 4) \times C(12, 7)$$
$$\text{Total ways} = 5 \times 792 = \mathbf{3960}$$
8. Selecting 2 black and 3 red balls from 5 black and 6 red balls.
The selections are independent.
- Select 2 black balls from 5 black balls:$$C(5, 2) = \frac{5 \times 4}{2 \times 1} = 10 \text{ ways}$$
- Select 3 red balls from 6 red balls:$$C(6, 3) = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \text{ ways}$$
$$\text{Total ways} = C(5, 2) \times C(6, 3)$$
$$\text{Total ways} = 10 \times 20 = \mathbf{200}$$
9. Choosing a program of 5 courses where 2 specific courses are compulsory.
Total courses available: 9.
Total courses to be chosen: 5.
Compulsory courses: 2.
Since 2 courses are compulsory, they must be included in the selection. We only need to choose the remaining $5 – 2 = 3$ courses from the non-compulsory courses.
Non-compulsory courses available: $9 – 2 = 7$ courses.
We select 3 courses from the 7 non-compulsory courses:
$$\text{Number of ways} = C(7, 3)$$
$$C(7, 3) = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 7 \times 5 = \mathbf{35}$$
