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I am given three natural numbers $n>x\geq k$ and trying to find the value of $$ \binom{n}{k}-\binom{n-x}{k} $$

I reached the following and not sure if I'm in the right path; $$ \frac{n!-(n-x)!\prod\limits_{i=1}{x}(n-x-k+i)}{k!(n-k)!\prod\limits_{i=1}{x}(n-x-k+i)} $$

I only need some hints so I can do it by myself. I also would like if someone give me a hint over the largest value of k (for a fixed $x$) where the result of subtraction will be more than the half of $\binom{n}{k}$.

seteropere
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  • Do you have reason to expect there to be a cleaner way to write the expression than simply $\binom{n}{k}-\binom{n-x}{k}$? That seems pretty clean as it is... – JMoravitz Nov 23 '16 at 02:11
  • @JMoravitz only my ignorance in such expressions. I felt there should be some closed form of the subtraction. – seteropere Nov 23 '16 at 05:49

1 Answers1

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It is already neat, but if you insist on changing it: $$ \binom{n}{k}=\binom{n-x+x}{k}=\frac{(n-x)!(n-x+1)\cdots n}{(n-x-k)!(n-x-k+1)...(n-k)k!}=\frac{n!}{(n-k)!k!} $$

and

$$ \binom{n-x}{k}=\frac{(n-x)!}{(n-x-k)!k!}\\ $$ Therefore

$$ \begin{align} \binom{n}{k}-\binom{n-x}{k}&=\frac{(n-x)!}{(n-x-k)!k!}\left(\frac{(n-x+1)\cdots n}{(n-x-k+1)...(n-k)}-1 \right)\\[10pt] &=\binom{n-x}{k}\left(\frac{P(n,x)}{P(n-k,x)}-1 \right) \end{align} $$

where $$P(n,k)=\frac{n!}{(n-k)!}$$

msm
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