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

Question

I have to simplify $\sum_{k=0}^n (-1)^k {{n}\choose{ k}} {{k}\choose{ j}}$. I found following identity which might be useful $(-1)^i{{x}\choose{ i}} = {{i - 1 - x}\choose{ i}}$ [but I don't know how it's possible since $i - 1 - x$ is negative]. Could you help me?

Answer 1

Let us demote the sum as $S_{n,j}$ and Use the property that $${n \choose k}{k \choose j}= {n \choose j} {n-j \choose k-j}.$$ Then $$S_{n,j}=\sum_{k=0}^{n} (-1)^k {n \choose k} {k \choose j}=\sum_{k=0}^{n} (-1)^k {n \choose j} {n-j \choose k-j}= {n \choose j} \sum_{k=0}^{n}(-1)^k {n-j \choose k-j}. $$ In $S_{n,j}$ above, let $k-j=p$, Then $$S_{n,j}=(-1)^j {n \choose j} \sum_{p=-j}^{n-j} (-1)^p {n-j \choose p}=(-1)^j {n \choose j} (1-1)^{n-j}= (-1)^n \delta_{n,j},$$ where $\delta_{n,j}$ is the Kronecker delta which is 1 if $j=n$ and zero otherwise.