So I'm trying to prove the following:

$$ \sum\limits_{k=0}^{n} {k \choose a} {n-k \choose b} = {n+1 \choose a+b+1}$$

And I'm a little caught on how to get started. It doesn't seem like a straightforward manipulation of the binomial theorem, so I was wondering if there was a nice way to visualize this in a combinatorial way -- i.e. I see that we'd be multiplying the ways of finding a subset of size a from a set of size $k$ and the ways of finding a subset of size $b$ from a set of size $n-k$, and this should equal the ways of finding a subset of size $a+b+1$ from a set of size $n+1$. I was wondering if this type of story could be extended to prove this, or if another approach was more advantageous.

This is a homework problem, tried to put as much of my thought process down as possible.


Suppose you have $n + 1$ positions and $a + b + 1$ markers. Put one marker in the $(k+1)^{\text{st}}$ position and place $a$ markers to the left and $b$ markers to the right (with no two markers occupying the same position). How many ways are there of doing this?

  • $\begingroup$ k is arbitrary so we'd have to do it for each k. But fixing one, you'd have (trying not to look at the left side as much as possible, I swear) $k \choose a$ markers (assuming you had a typo) times $n-k \choose b$ markers. $\endgroup$ – Schwinger Sep 23 '14 at 4:05
  • $\begingroup$ Not $\binom{k}{a}$ markers, but $\binom{k}{a}$ ways of choosing the $a$ markers. Can you see how this helps? $\endgroup$ – Michael Albanese Sep 23 '14 at 4:09
  • $\begingroup$ Right, ways, and yes - thanks! $\endgroup$ – Schwinger Sep 23 '14 at 4:12
  • $\begingroup$ @user178092: If this answer was useful you should upvote it (click the up arrow next to the answer). If this answer has helped you to solve your problem, you can also accept it (click the tick next to the answer). $\endgroup$ – Michael Albanese Sep 23 '14 at 23:44

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