TITLE: Why is this sum equal to 1?
QUESTION [0 upvotes]: I'm sorry for opening a new question for a triviality, but I couldn't find this result online.
The following holds:
$$\sum_{j=0}^{n-1} {\dfrac{ {{K-1}\choose{j}} {{N-1-(K-1)}\choose{n-1-j}} } {{N-1}\choose{n-1} }}=1$$
It looks like an identity that I'm not aware of. I checked with easy numbers and all terms are different than 0. I only know of this identity $\sum_{k=0}^n {{n}\choose{k}}=2^n$ which I don't think is any helpful.
Both $-1$ in the second binomial in the numerators have been added from a previous step, I guess in order to make it easier to understand why the sum is actually $1$.
I'd be glad if you can help me. Thank you.

REPLY [1 votes]: Observe that by the 1 Chu-Vandemonde Identity
$$\sum_{j=0}^{n-1}\binom{k-1}j\binom{N-k}{n-1-j}= \binom{N-1}{n-1}$$
...and we're done
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