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

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 combinatorical 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!

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 $$\binom{2k}{k-1}$$

• +1 Actually there is no harm in allowing $r=0$, as there is none in allowing $r=k+1$, or even in summing over all $r\in\mathbf Z$ to make clear that the range of summation is not producing a "cut-off" effect that would spoil using the Vandermonde convolution – Marc van Leeuwen Apr 13 '13 at 7:40


• You really like that formula that uses the contour integral, don't you? ;) (You use it a lot...) – apnorton Dec 2 '14 at 16:53
• @anorton Yes. I really like it because it leads to straightforward calculations. However, it doesn't give any insight toward the 'counting or/and combinatoric' procedure. Thanks. – Felix Marin Dec 2 '14 at 18:35