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Let $C(n,r)$ denote the number of ways of choosing $r$ objects from $n$ objects (order does not matter). I want to prove that for any positive integers $u>m$, $$ \sum_{r=m}^u (-1)^{r-m}C(r,m)C(u,r)=0 $$

I tried to use the binomial formula to write the left-hand-side as a power of $0$, but it does not work.

Tony B
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    Try using the fact that $\binom{u}{r}\binom{r}{m}=\binom{u}{m}\binom{u-m}{r-m}$ and then using your idea. Notice that $\binom{u}{m}$ has nothing to do with the indexing variable and can be factored outside. – JMoravitz Sep 26 '17 at 22:40

2 Answers2

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Let $k=r-m$. We have then:

$\sum\limits_{r=m}^u(-1)^{r-m}\binom{r}{m}\binom{u}{r} = \sum\limits_{r=m}^u(-1)^{r-m}\binom{u}{m}\binom{u-m}{r-m}$

$=\binom{u}{m}\sum\limits_{r=m}^u(-1)^{r-m}\binom{u-m}{r-m}$

$=\binom{u}{m}\sum\limits_{k=0}^{u-m}(-1)^k\binom{u-m}{k}$

$=\binom{u}{m}\sum\limits_{k=0}^{u-m}(-1)^k(1)^{u-m-k}\binom{u-m}{k}$

$=\binom{u}{m}(1-1)^{u-m}$

$=\binom{u}{m}(0)^{u-m}$

$=0$

Notice the last line can be explained since $u>m$ we have $0$ to the power of a nonzero number. In the case that $u=m$ this step would have been invalid.

JMoravitz
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$(-1)^mC(r,m)$ is a polynomial in $r$ of degree $m$. It is zero for $r<m$, so we can write the sum as $$ \sum_{r=0}^u (-1)^r C(u,r) P(r), $$ where $P(r)=(-1)^mC(r,m)$ is that polynomial. Then one can use Euler's result on differences of $k$th powers to conclude that this is zero since $m<u$. (So that the sum contains a $(-1)^mC(r,m)$ is not really significant itself: the significant part is that it's a polynomial).

Chappers
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