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I'm required to prove the following binomial identity:

$$\sum\limits_{k=0}^l {n \choose k} {m \choose l-k} = {n+m \choose l}$$

I tried various arrangements but reached nowhere. Finally I turned to the hint in the book, which says

Apply the binomial theorem to $(1+x)^n (1+x)^m$

And suddenly, it makes sense. All I now need to do is add the powers on the right-hand side and equate the coefficients of $x^l$. But I'm wondering how to write a proper proof. Will it be enough if I say:

$(1+x)^n (1+x)^m = (1+x)^{m+n}$

Applying the binomial theorem separately for the two terms on the LHS and collecting the coefficients of $x^l$ on both sides, we have:

$${n \choose 0} {m \choose l} + {n \choose 1} {m \choose l-1} + \ldots + {n \choose l}{m \choose 0} = {n+m \choose l}$$

Is this enough? I don't know why but it looks rather shallow to me.

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    $\begingroup$ Your proof is correct. I am not sure what is bothering you. If you want to prove that necessarily the coefficients of $x^l$ are equal, you can note that both the LHS and RHS are polynomials of the variable $x$, therefore they coincide on $\mathbb{R}$ if and only if they have the same coefficients. $\endgroup$ – Ian May 9 '14 at 4:37
  • $\begingroup$ @Ian: Thanks a lot! You know how it is ... when a proof is so short and "obvious" you start wondering if it's correct. ;) $\endgroup$ – dotslash May 9 '14 at 4:38
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    $\begingroup$ yes it is right.One should always be able to do such easy sums.I believe. @ dotslash $\endgroup$ – soumajit das May 9 '14 at 4:52
  • $\begingroup$ There are several posts about this identity: See this question and other posts, which are linked there. $\endgroup$ – Martin Sleziak May 10 '14 at 16:23
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Other proof is counting in two ways: what the number of choose $l$ balls in $m+n$? (the balls are enumerate -- 1,2,...,m+n)

for one side: $\binom{m+n}{l}$

for other side: divided the balls into 2 groups, group 1 with $n$ balls and group 2 with $m$ balls. If we choose $k$ balls in group 1, we should choose $l-k$ balls in group 2. So, we have $\sum\limits_{k=0}^l {n \choose k} {m \choose l-k} = {n+m \choose l}$

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  • $\begingroup$ Very nice answer, I must say! Combinatorial proofs are so much more interesting. $\endgroup$ – dotslash May 10 '14 at 17:14
  • $\begingroup$ thank you :) I like combinatorial proofs too! $\endgroup$ – leticia May 10 '14 at 17:18
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I'm answering my own question so that it can be closed. As explained by @Ian in the comments above:

"Your proof is correct. I am not sure what is bothering you. If you want to prove that necessarily the coefficients of xl are equal, you can note that both the LHS and RHS are polynomials of the variable x, therefore they coincide on R if and only if they have the same coefficients."

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