How do I simplify factorials?

Question

Given the following equation:

$$\frac{n!}{(n-k)!k!}*\frac{1}{n^{k}}=\frac{1}{k!}*\frac{n*(n-1)...(n-k+1)}{n^{k}}$$

(I used * as the multiplication sign).

I understand $\frac{1}{k!}$ but $$\frac{n*(n-1)*...*(n-k+1)}{n^{k}}$$ is a problem for me. I also understand why $n^{k}$ is in the denominator. I specifically do not understand what happened to $n!$ and $(n-k)!$

Answer 1

I think what you mean is $$ \frac{n!}{(n-k)!} = \frac{1 \times 2 \times \ldots \times n}{1 \times 2 \times \ldots \times (n-k)} = n (n-1)\ldots(n-k+1) = P(n, k) $$ Also note that this product can bounded: $$ P(n,k) = n^k \times (1-\frac{1}{n})(1-\frac{2}{n}) \ldots (1-\frac{k-1}{n}) \leq n^k $$

Answer 2

Notice if $m > n$ then $\frac {m!}{n!} = \frac {1*2*3*4*........*n*(n+1)*......m}{1*2*3*4*......*n} = \frac {(1*2*3*4*......*n)*[(n+1)*.....m])}{1*2*3*4*......*n}=(n+1)*.....m$

So $\frac{n!}{(n-k)!k!}*\frac{1}{n^{k}}=$

$\frac 1{k!}*\frac 1{n^k} * \frac {n!}{(n-k)!} =\frac 1{k!}*\frac 1{n^k} *\frac {[1*2*3 *......*(n-k)]*[(n-k+1)*(n-k+2)*......*n]}{1*2*3 *......*(n-k)}=\frac 1{k!}*\frac 1{n^k}(n-k+1)*(n-k+2)*......*n$