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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.

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    $\begingroup$ Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. $\endgroup$
    – Shaun
    Feb 13 '21 at 1:52
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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.

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  • $\begingroup$ Why assume k = 1? Why not for m ? $\endgroup$
    – S.M.T
    Feb 13 '21 at 4:11
  • $\begingroup$ I am not assuming $k=1$, $k\ge0$ (I just stated the division of the result for k=0). However, P(N,S) is always divisible by P(N,R) given the condition $N\ge S\ge R$ and the result of the division depends on k and m. $\endgroup$
    – tempx
    Feb 13 '21 at 12:52
  • $\begingroup$ I meant to write why assume k =0.actually. Why only stated the result of division for k =0. Does that on,y prove the answer ? Why not other situations $\endgroup$
    – S.M.T
    Feb 13 '21 at 13:48
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    $\begingroup$ oh ok, I have stated $k=0$ specifically since in my answer I have written $\frac{(k+m)!}{m!}=(m+1)(m+2)...(m+k)$ where $(m+1)(m+2)...(m+k)$ doesn't include the case for $k=0$. If $k=1$, then the answer is $m+1$, if $k=2$, the answer is $(m+1)(m+2)$ and so on. That is why I specifically stated the result for the division for the case $k=0$ which is 1. $\endgroup$
    – tempx
    Feb 13 '21 at 14:57
  • $\begingroup$ thanks I got it . $\endgroup$
    – S.M.T
    Feb 13 '21 at 16:19
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Start by proving that the quotient $\frac{P(n,s)}{P(n,r)}$ is indeed an integer.

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  • $\begingroup$ Ok but what’s the reason for that ? $\endgroup$
    – S.M.T
    Feb 13 '21 at 4:12

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