The question is as follows, I think I solved it partially:
Show that there are $a,b$ real positive numbers such that
$an^7 \leq \frac{n!}{7!(n-7)!} \leq bn^7$
$7\leq n$
my solution for b:
$\frac{n!}{7!(n-7)!} \leq \frac{n!}{(n-7)!} = (n-6)(n-5)(n-4)(n-3)(n-2)(n-1)(n) \leq n*n*n*n*n*n*n = n^7$ so $b=1$ is a good answer for b.
But for a...I'm stumped
For $n\ge7$, $n-k\ge n/7$ for $0\le k\le6$
So $$\frac{n!}{7!(n-7)!} =\frac{(n-6)(n-5)(n-4)(n-3)(n-2)(n-1)n}{7!}\ge \frac{n^7}{7!7^7}$$
$$a=\frac{1}{7!7^7}$$