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I am trying to prove the equation below with P.I.E :

$$\sum_{i=0}^{n}(-1)^n\binom{n}{i}\binom{m+n-i}{k-i} = \binom{m}{k}$$

First RHS is quite simple, i.e., choosing k among m, and then move to LHS, this part is little confusing to me.

It represents the task of applying P.I.E. through the part $\binom{n}{i}\binom{m+n-i}{k-i}$, however, with analogy to the latter part of P.I.E,

$$\sum_{1\le i_1\lt i_2\cdot\cdot\cdot\lt i_k \le n } \mid A_{i_1}\cap\cdot\cdot\cdot \cap A_{i_k} \mid$$

I can't understand how I can link this part to the given product of binoms.

any guidance to the direction?

Quanlisp
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1 Answers1

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$Z = X\cup Y$, where $X =\{x_{1} , . . . , x_{n}\}$ is an $n$-set of blue points and $Y$ is an $m$-set of red points. How many $k$-subsets consist of red points only? Clearly its $\binom{m}{k}$ which is your right hand side. Now to count the left hand side take $S$ to be all the $k$-subsets of $Z$ and $Z_{i}$ to be those $k$-subsets that contain $x_{i}$ and then apply inclusion-exclusion.

C.S.
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