A combinatorial argument that $5$ divides $(n+2)(n+1)n(n-1)(n-2)$.

Question

Give a combinatorial argument to prove that $5$ divides $(n+2)(n+1)n(n-1)(n-2)$.

I am definitly weakest in discrete math, but it jumps out to me to use the permutation definition that

$$P(n,m)=n\cdot(n-1)\cdot(n-2)\cdots (n-(m-1))=\frac{n!}{(n-m)!}$$

Pretty clearly here $$P(n+2,5)=(n+2)(n+1)n(n-1)(n-2)=\frac{(n+2)!}{(n+2-5)!} $$ Where I seem to want to go from here is that $P(m,5)$ is always divisible by $5$. Heres the 'proof' that I feel a bit shakey on.

Let $m$ be arbitrary, then $$P(m,5)=\frac{m!}{(m-5)!}$$ We require $m-5\geq0$, so $m\geq5$. This means $m!$ must be divisible by $5$ and hence $P(m,5)$ must also be divisible by $5$.

Answer 1

Combinatorial argument:

$N=(n+2)(n+1)n(n-1)(n-2)$ represents the number of ways you can create a row of 5 people from a group of $n+2$ people (there are $n+2$ ways to select the first one, $n+1$ to select the second one, etc...)

Say that one row of people looks like $ABCDE$. How many rows do you have with the same set of people? It's 5! (one of five guys ABCDE can be selected to be first, one of the remaining four to be the second, etc...). So $N$ must be divisible by 5! which also means that $N$ must be divisble by 5.

Answer 2

$$(n+2)(n+1)n(n-1)(n-2)=5!\binom{n+2}{5}$$ and since $\binom{n+2}{5}$ is a natural number for $n\geq 3,$ we see that $(n+2)(n+1)n(n-1)(n-2)$ is divisible by $5$.