WAEC 2013 · Paper 2 · Q6

(a)

1. Use ${}^nC_4 = \dfrac{n(n - 1)(n - 2)(n - 3)}{4!}$, and $4! = 24$.
2. So ${}^{p+1}C_4 = \dfrac{(p + 1)p(p - 1)(p - 2)}{24}$.
3. And ${}^{p-1}C_4 = \dfrac{(p - 1)(p - 2)(p - 3)(p - 4)}{24}$.
4. Take out the common factor $\dfrac{(p - 1)(p - 2)}{24}$: the difference is $\dfrac{(p - 1)(p - 2)}{24}\left[(p + 1)p - (p - 3)(p - 4)\right]$.
5. Expand the bracket: $p^2 + p - (p^2 - 7p + 12) = 8p - 12$.
6. Write $8p - 12$ as $4(2p - 3)$ and cancel 4 into 24.
7. So ${}^{p+1}C_4 - {}^{p-1}C_4 = \dfrac{(p - 1)(p - 2)(2p - 3)}{6}$.