I have a problem which requires multiplying:
$$ \frac 1 4 \cdot \binom n k $$
Expanded: $$ \frac 1 4 \cdot \frac {(n)!} {5!(n-5)!} $$
The answer is below, but it isn't clear how to get from the initial state to this one algebraically. $$ \frac {(n-1)!} {4!(n-5)!} $$
If $n=20$ then $$\frac{1}{4}\cdot\frac{n!}{5!\cdot(n-5)!}=\frac{1}{4}\cdot\frac{20!}{5!\cdot15!}=\frac{20\cdot19!}{4\cdot5\cdot4!\cdot15!}=\frac{19!}{4!\cdot15!}=\frac{(n-1)!}{4!\cdot(n-5)!}$$