Binomial coefficient sum: $\sum_{k=0}^{i-1}{i-1 \choose k}{j-1\choose k} = {i+j-2\choose j-1}$

Question

I'm having problems showing this equation, hope someone here can help me:

$$\sum_{k=0}^{i-1}{i-1 \choose k}{j-1\choose k} = {i+j-2\choose j-1}$$

where $1 \leq i\leq j $.

well the problem stems from trying to show the decomposition of a matrix, but you're right. — J. Doe, Jun 17 '20 at 10:19
See also: Combinatorial interpretation for the identity $\sum\limits_i\binom{m}{i}\binom{n}{j-i}=\binom{m+n}{j}$?, Evaluate $\sum_{k=0}^{n} {n \choose k}{m \choose k}$ for a given $n$ and $m$. (and other posts linked there). — Martin Sleziak, Jun 17 '20 at 15:20
Once it is noted that $\binom{j-1}{k}=\binom{j-1}{j-k-1}$, this is known as Vandermonde's Identity. — robjohn, Jun 17 '20 at 15:26

Benjamin Wang · Accepted Answer · 2020-06-17T14:25:43.373

1

Note that $\binom{j-1}{k} = \binom{j-1}{j-1-k}.$

Then both sides are different ways of counting how to take $j-1$ elements from $\{1, 2, \ldots , i+j-2\}$.

Edit to clarify:

Imagine there are $i-1$ boys and $j-1$ girls, and you want to form a committee of $j-1$ people. How many ways to do this are there? Obviously, one way of expressing this number is the RHS. On the other hand, you can pick $k$ boys first, then pick $j-1-k$ girls. This gives one term on the LHS. We sum this for all possible values of $k$ to give the LHS. (Note: the expression works if there are at least as many girls as boys. Otherwise, the range of $k$ will overshoot.)

edited Jun 17 '20 at 14:25

answered Jun 17 '20 at 10:14

Benjamin Wang

3,920

sorry I don't see why that is true for the LHS.. – J. Doe Jun 17 '20 at 10:30
I think this overcounts,because, e.g. for $k=1 \ \binom{i+j-2}{2} = \binom{i-1}{1} + \binom{j-1}{1} + \binom{i-1}{2}+\binom{j-1}{2}$ – Alex Jun 17 '20 at 10:31
Question Asker, I've edited to clarify. Does this make sense? @Alex I'm not sure what you're trying to say. Sorry. – Benjamin Wang Jun 17 '20 at 14:26
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Thanks i understand now. This is a Vandermonde identity, right? – J. Doe Jun 17 '20 at 15:19
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@J.Doe: Yes, it’s the Vandermonde identity. – Brian M. Scott Jun 17 '20 at 15:20

Binomial coefficient sum: $\sum_{k=0}^{i-1}{i-1 \choose k}{j-1\choose k} = {i+j-2\choose j-1}$

1 Answers1