5

I have to prove the equality $$\binom{n}{2k+1}=\sum_{i=1}^n{\binom{i-1}{k}\binom{n-i}{k}}$$

What I can see is that the left hand side is the number of ways to choose $2k+1$ elements from $n$ elements, while the right hand side is the sum of the ways you can choose k elements from one set, and k from another, effectively choosing $2k$ elements from a set of $n-1$ elements.

This doesn't make sense to me. I would expect that the left hand side would be the sum of the ways to choose $2k+1$ elements from $n$ elements, choosing from two different sets.

Can anyone help me with this?

gsgx
  • 789
  • 1
    To choose $2k+1$ elements randomly you could first pick $i$ to be the median and then from each side of $i$ choose additional $k$ elements. – dtldarek Mar 09 '12 at 21:08
  • See http://math.stackexchange.com/questions/98495/formula-for-binomial-coefficients

    http://math.stackexchange.com/questions/73015/proof-of-sum-0-le-k-le-t-t-k-choose-rk-choose-s-t1-choose-rs1

     –  Mar 09 '12 at 21:19

2 Answers2

5

Hint: $$\sum_{i=1}^n{\binom{i-1}{k}\binom{n-i}{k}}=\sum_{i=1}^n{\binom{i-1}{k}\binom{n-i}{k}}\binom{1}{1}$$

Daniel Montealegre
  • 6,783
  • I upvoted at first, but this is irrelevant. It does not reflect the combinatoric argument behind the identity. See my answer. You should consider editing your answer to make it better or remove it. – Patrick Da Silva Mar 09 '12 at 20:46
  • 4
    @Patrick: How is it irrelevant? Of course your explanation is more elaborate, but it essentially comes down to dividing the set in three parts of size $i-1, 1, n-i$, and picking $k, 1, k$ in each of them. – TMM Mar 09 '12 at 20:54
  • Because it does not suggest ordering, of course we know that there is one last element to be chosen to form the set of $2k+1$ elements, the whole problem is to how to get it. – Patrick Da Silva Mar 09 '12 at 21:02
  • 1
    One way to make it relevant would be this :

    HINT : $$ \sum_{i=1}^n \begin{pmatrix} i-1 \ k \end{pmatrix}\begin{pmatrix} n-i \ k \end{pmatrix} = \sum_{i=1}^n \begin{pmatrix} i-1 \ k \end{pmatrix}\begin{pmatrix} 1 \ 1 \end{pmatrix}\begin{pmatrix} n-i \ k \end{pmatrix} $$ It seems stupid, but it suggests to place the $i^{\text{th}}$ element in the middle, which is a way more useful hint than the one you suggested, which just suggests to "choose one more to complete the set".

     – Patrick Da Silva Mar 09 '12 at 21:04
  • 6
    Placing that extra binomial in the middle may already qualify as a full solution to some :) (But I agree that that would be even clearer, and the summation on the right does need interpretation to make the proof complete. I just wanted to say that I thought the "irrelevant" in your comment was a bit harsh.) – TMM Mar 09 '12 at 21:13
  • 2
    @Patrick: Is not completely irrelevant. It was just a hint, by no means it was an attempt of a solution. I will not edit it, make it better, nor remove it, because it meant for him to realize that he was "forgetting" about another element when he was doing his double counting argument. – Daniel Montealegre Mar 09 '12 at 21:18
  • @TMM : Yeah, maybe a little bit harsh... I'd edit my comment but I can't >.> let's say I wrote "not very helpful" instead of "irrelevant". – Patrick Da Silva Mar 09 '12 at 21:28
5

Enumerate your $n$ elements as $\{ 1,2,\dots, n\}$. We will count the number of ways to select $2k+1$ integers amongst them. To do so, select the $k+1$ first ones. The $k+1^{\text{th}}$ number will be at the $i^{\text{th}}$ position, and then select the $k$ last ones. Fixing the position of the $i^{\text{th}}$ number, there are $\begin{pmatrix} i-1 \\ k \end{pmatrix}$ possibilities for the $k$ first ones and $\begin{pmatrix} n-i \\ k \end{pmatrix}$ for the last $k$ ones. You do not double count/miss any possibility because we fixed an order of the elements of your set, and if the $i^{\text{th}}$ element is fixed, only one possibility picks the $k$ ones chosen before the $i^{\text{th}}$ and and the $k$ ones after it. Summing over $i$ gives you the result.

Hope that helps,

Patrick Da Silva
  • 41,413