$\sum_{r=0}^{n-k} {n \choose r}{n \choose {r+k}}={n \choose {n-k}}$

Question

$$\sum_{r=0}^{n-k} {n \choose r}{n \choose {r+k}}={n \choose {n-k}}$$

Can you prove why this result is as such? I have no idea how to start proving this property.

Thank You!

Answer 1

If I take $n = 3$ and $k=1$ then I have : $$\sum_{r=0}^{2} {3 \choose r}{3 \choose {r+1}}= {3 \choose 0}{3 \choose {1}} +{3 \choose 1}{3 \choose {2}} + {3 \choose 2}{3 \choose {3}} = 3 + 9 + 3 = 15 \neq {3 \choose {2}} = 3$$

Therefore it is false.

Otherwise, it seems very similar to the Vandermonde's identity that you can prove with Newton's binomial.

I think you meant :

$$\sum_{r=0}^{n-k} {n \choose r}{n \choose {r+k}}={2n \choose {n-k}}$$