WAEC 2009 · Paper 2 · Q13

(a)
If ${}^nP_3 = 10\left({}^{n-2}P_2\right)$ , find $n$ .
Separate values with commas, e.g. 3, −2
(b)(i)
Four students are to be selected from 4 boys and 6 girls to represent a school in a competition. If there is no restriction, in how many ways can they be selected?
(b)(ii)
Calculate the probability of having an equal number of boys and girls.

Worked solution (try it first)

(a)

${}^nP_3 = \dfrac{n!}{(n - 3)!}$
$= n(n - 1)(n - 2)$ and ${}^{n-2}P_2 = \dfrac{(n - 2)!}{(n - 4)!}$
$= (n - 2)(n - 3)$ .
So $n(n - 1)(n - 2) = 10(n - 2)(n - 3)$ .
Divide by $n - 2$ (it isn't zero, since $n \ge 4$ ): $n(n - 1) = 10(n - 3)$ .
Expand: $n^2 - n = 10n - 30$ , so $n^2 - 11n + 30 = 0$ .
Factorise: $(n - 5)(n - 6) = 0$ , so $n = 5$ or $n = 6$ .
Both work: ${}^5P_3 = 60 = 10 \times {}^3P_2$ and ${}^6P_3 = 120 = 10 \times {}^4P_2$ .

(b)(i)

Order doesn't matter, so choose 4 of the 10 students: ${}^{10}C_4 = 210$ ways.

(ii)

Equal numbers means 2 boys and 2 girls: ${}^4C_2 \times {}^6C_2 = 6 \times 15 = 90$ ways.
Probability $= \dfrac{90}{210} = \dfrac37$ .

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