Q.Simplify $\dfrac{(n+2)!}{n!} - \dfrac{n!}{(n-1)!}$, and hence evaluate it for $n=5$.

Each ratio of factorials collapses to a short polynomial once the common factorial in numerator and denominator is cancelled, which is the whole point of factorial notation.

Recall that $n! = n(n-1)(n-2)\cdots1$, so a ratio of factorials cancels all but the extra leading terms.

$\dfrac{(n+2)!}{n!} = \dfrac{(n+2)(n+1)\,n!}{n!} = (n+2)(n+1)$.

$\dfrac{n!}{(n-1)!} = \dfrac{n\,(n-1)!}{(n-1)!} = n$.

So the expression becomes $(n+2)(n+1) - n = (n^2+3n+2) - n = n^2+2n+2$.

At $n=5$: $5^2+2(5)+2 = 25+10+2 = 37$. …

Cancelling factorials incorrectly, e.g. writing $\dfrac{(n+2)!}{n!}$ as simply $(n+2)$ instead of $(n+2)(n+1)$ — every factor strictly bet …