-1
$\begingroup$

I'm trying to prove the following: $$\binom{n + p}{k} = \sum_{j=0}^n \binom{n}{j} \cdot \binom{p}{k - j}$$ How do I do it? Induction? And can someone hint me at how to start?

$\endgroup$
2
  • 1
    $\begingroup$ Consider the coefficient of $x^k$ in $(1+x)^{n+p}$ as well as in $(1+x)^n(1+x)^p$. $\endgroup$
    – Kenny Lau
    May 7, 2016 at 14:34
  • $\begingroup$ Several proofs here: Vandermonde's identity $\endgroup$ May 9, 2016 at 5:45

1 Answer 1

0
$\begingroup$

In totaly, you have $n+p$ items, you keep $n$ items in box 1 and keep $p$ items in box 2.

You have to pick a total of $k$ items.

If you pick $j$ items from box 1, you have to pick $k-j$ items from box 2 to make it up.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged .