JAMB 2003 · UME · Q47

Write both in terms of $n$: $^nP_3 = n(n - 1)(n - 2)$ and $6\,(^nC_4) = \dfrac{6\,n(n - 1)(n - 2)(n - 3)}{24}$

Set them equal and divide both sides by $n(n - 1)(n - 2)$, which is not zero: $1 = \dfrac{n - 3}{4}$.

So $n - 3 = 4$ and $n = 7$, option D.