WAEC 2023 · Paper 2 · Q1

Permutation & combination

Given that ${}^nC_2 = 66$ , find the value of $n$ .

(a)
Value of $n$

Worked solution (try it first)

$\,{}^nC_2 = \dfrac{n(n - 1)}{2} = 66$ , so $n(n - 1) = 132$ .
Rearrange: $n^2 - n - 132 = 0$ .
Factorise: $(n - 12)(n + 11) = 0$ .
$n$ is a positive whole number, so $n = 12$ .

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