Question: prove that if $R\leq S$ and $S\leq N$ then $P(N,S)$ is divisible by $P(N,R)$

Let $s = r +k$ since s is greater than r by some value but we don’t we don’t how much so I used k

$\frac{n!}{(n-(r+k))!}$/$\frac{n}{n-r}!$

$\frac{n!}{(n-r-k))!}$/$\frac{n}{n-r}!$ where you can cancel out n!

I am not getting how to solve farther than this.

Now , I also don’t the meaning behind this question . Because $r$ can be just any value. There may be many values of $P(N,S)$ divisible by $P(N,R)$. Many different values for $N , R$ and $S$. so , from this way of solving . Can we find at least which value is $n,r$ and $s$ also. Because i think just by dividing variables , how can I find the real values when they are just not given.

Answer: 
I believe you are so close to the solution. As you suggested let

$$
s=r+k
\\n=s+m=r+k+m
$$

where $k\ge0$ and $m\ge0$. Then

$$
\frac{P(N,S)}{P(N,R)}=\frac{n!}{(n-s)!}\frac{(n-r)!}{n!}=\frac{(k+m)!}{m!}=(m+1)(m+2)...(m+k)
$$

if $k=0$, then the division becomes 1.
Start by proving that the quotient
$\frac{P(n,s)}{P(n,r)}$
is indeed an integer.