# Combinatorial proof of $\sum_{j=i}^n { n \choose j} { j \choose i} (-1)^{n-j}=0$

I would like a combinatorial proof using committee and subcommittee selection for the following identity.

$$\sum_{j=i}^n { n \choose j} { j \choose i} (-1)^{n-j}=0$$

with $i<n$.

This is from Ross's A First Course in Probability, problem number 14c.

I see it this way:

$${ n \choose j} { j \choose i} = { n \choose i} { n- i \choose j-i}$$

So the sum becomes

$$\sum_{j=i}^n { n \choose i} { n- i \choose j-i} (-1)^{n-j} = { n \choose i} \sum_{j=i}^n { n- i \choose j-i} (-1)^{n-j}$$

The right hand side becomes ${ n \choose i} \sum_{k=0}^{n-i} { n- i \choose k} (-1)^{n-i-k}$. Since $\sum_{k=0}^{n-i} { n- i \choose k} (-1)^{n-i-k} =0$. The identity hold.

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$$\binom{n}{j}\binom{j}{i}=\binom{n}{n-j}\binom{j}{j-i}$$