Finding a closed formula for $\sum_{r=0}^k {k \choose r}{k \choose r-1}$

Question

I'm looking to find a closed formula for: $\sum_{r=0}^k {k \choose r}{k \choose (r-1)}$. My conjecture is that it is equal to ${2k \choose k-1}$ but I could not prove it. I'm interested in knowing if there's a way to prove this identity (preferably a combinatorial one), but really any would do. my preference is due to the fact that it helps understanding the intuition behind the proof. Thanks a million!

Answer 1

Your conjecture is correct (though the sum probably needs to start with $r=1$).

You can rewrite as

$$\sum_{r=1}^{k} \binom{k}{k-r} \binom{k}{r-1}$$

which has a combinatorial interpretation.

Picking $k-1$ things from $k+k$ things (choose $r-1$ from first $k$, remaining from the second), which is $$\binom{2k}{k-1}$$