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My question is:

show $$\sum_{k=0}^{m} \binom{n-k}{m-k}=\binom{n+1}{m}$$ $$n\geq m\geq 1$$ I tried to do this via induction and failed. there has to be another way of doing this.

We could either explain combinatorically that we are counting the same thing in 2 different ways, in this sense, we are counting the number of ways to choose $m$ elements from a set with $n+1$ elements.

Or we could try like me to prove it algebraically, both ways are valid...I'd like some help, or a push in the right direction.

user127.0.0.1
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Oria Gruber
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    maybe his could help

    $\binom{n+1}{m}=\binom{n}{m}+\binom{n}{m-1}$

     – derivative Jan 19 '14 at 20:04
  • If I am not mistaken, this is (in a different form) the same identity as in the following questions: http://math.stackexchange.com/questions/74844/induction-proof-concerning-a-sum-of-binomial-coefficients http://math.stackexchange.com/questions/544503/combinatorial-approach-to-sum-limits-i-1n-binomir-1i http://math.stackexchange.com/questions/205031/trying-to-prove-that-p-prime-divides-binomp-1k-binomp-2k-1-cdo – Martin Sleziak Jan 24 '14 at 12:42

2 Answers2

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Hint: The right side counts the number of ways to pick $m$ elements out of $n+1$.

For the left: partition the $m$-element subsets of $n+1$ based on the highest element that is NOT included in the subset.

Nick Peterson
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  • I'm not sure I understood you. You want me to partition the big set (of $n+1$ elements) to different sets of $m$ elements based on the highest element that isn't in the subset? I don't follow. – Oria Gruber Jan 19 '14 at 19:51
  • @OriaGruber Not at all. I want you to partition the collection of $m$-element subsets of ${1,2,\ldots,n+1}$ based on the highest absent element. Then the size of the full collection is the sum of the size of these parts... – Nick Peterson Jan 19 '14 at 19:54
  • if the highest element that is not used is $n-k$, then all the elements ${n-k+1,...n+1}$ are used. That's $k+1$ elements. that means that from the set ${1,2,...n-k-1}$ we use only $m-(k+1)$ elements – Oria Gruber Jan 19 '14 at 19:57
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    @OriaGruber Right. Now, to fix things a little bit, say that the highest element not used is $n+1-k$. Then $k$ is between $0$ (meaning that we don't use element $n+1$) and $m$ (meaning that we use all $m$ elements greater than $n+1-m$, and no others). For a given value of $k$, we use elements $n+1-k+1,\ldots,n+1$ ($k$ total), don't use element $n+1-k$, and have to use $m-k$ from ${1,2,\ldots,n-k}$. So... – Nick Peterson Jan 19 '14 at 20:00
  • yeah I understand now :) if $n-k$ is the largest value not used, that means ${n-k+1,...,n+1}$ are all used and there's no choice involved. However, now we need to choose $m-(k+1)$ elements from ${1,2,...,n-k-1}$ and for that I have $\binom{n-k-1}{m-k-1}$ options. since we took arbitrary such highest absent value, the sum of all of those (for all absent values) is the number of ways to choose m out of n+1. – Oria Gruber Jan 19 '14 at 20:03
  • Thanks for helping me with this. Just 2 last questions for you :D how did you know what the left side represents? Because I have more questions like this, I wont come here for help every time. And is there an algebraic solution? – Oria Gruber Jan 19 '14 at 20:06
  • @OriaGruber The big thought is this: on the right side, for each $k$, we've reduced the problem of choosing $m$ out of $n+1$ to the problem of choosing $m-k$ out of $n-k$. So, the question is really: how can you break the problem apart (so that there's no overlap!) in such a way that given $k$, we already know that exactly $k$ out of a given set of $k+1$ elements are used? I'm sure there's an algebraic solution, but I think that combinatorial arguments are much more instructive. – Nick Peterson Jan 19 '14 at 20:12
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As has already been suggested, this is nothing else than Pascal's rule applied recursively m times:

$${n+1\choose m}=\underbrace{n\choose m}_{k=0}+{n\choose m-1}=\underbrace{{n\choose m}+{n-1\choose m-1}}_{\begin{align}\text{The first terms of our series,}\\\text{for k=0 and k=1}\quad\end{align}}+{n-1\choose m-2}=\ldots=\sum_{k=0}^m{n-k\choose m-k}$$

Lucian
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