How to solve this 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.

• 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. Feb 13 '21 at 1:52

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.

• Why assume k = 1? Why not for m ? Feb 13 '21 at 4:11
• 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. Feb 13 '21 at 12:52
• 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 Feb 13 '21 at 13:48
• 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. Feb 13 '21 at 14:57
• thanks I got it . Feb 13 '21 at 16:19

Start by proving that the quotient $$\frac{P(n,s)}{P(n,r)}$$ is indeed an integer.

• Ok but what’s the reason for that ? Feb 13 '21 at 4:12