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:
$$\text{Number of ways} = C(7, 3)$$
$$C(7, 3) = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 7 \times 5 = \mathbf{35}$$