Solve the following equation algebraically for n. Show your working using factorial notation.? 

a) $\dfrac{n!}{(n-2)!} = n(n-1)$
so the expression equivalent to $$\dfrac{n(n-1)}{6} = 12 \implies n(n-1)=72$$
I think its obvious the required solution is $8 \cdot 9 =72$, so $n=9$
b) $8Pn = \dfrac{8!}{(8-n)!} = 6720$
$$(8-n)! = \frac{8!}{6720} = 6 = 3!$$
so $8-n =3 \iff n=5$

Looks good. Your answers are correct! Nice work.
You could also algebraically solve for $n$ in $(a)$, though I agree the solution seems "obvious": $$n(n-1) = 72 \iff n^2 - n - 72 = 0 $$ $$\iff (n-9)(n+8) = 0 \iff n = 9, n=-8$$
But since we are talking about non negative integers $n$ in these equations (since we are evaluating expressions with factorials), we can "throw out" $\;n = -8\,$ as an option, and we are left with $\;n = 9.$
