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i have to show that $\sum_{j=0}^n \binom{n}{j} * \binom{m}{k-j} = \binom{n+m}{k}$ is valid. I tried to do it by induction, the induction beginning fits, but unfortunately, I don't know how to the induction step. I have been failing now for hours, is there someone who can help me, please?

SR23
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  • I'm sure there are already several identical posts. Searching now for the closest fit, but the combinatorial proof is that: If you wish to pick $k$ people out of $(n+m)$ people (order not mattering) and those $(n+m)$ people can be separated into two groups, one of size $n$ and the other of size $m$, you can either pick the $k$ of them out of the group as a whole for $\binom{n+m}{k}$ possibilities, or you may choose instead to pick $j$ of them specifically out of the first group, and the $k-j$ remaining out of the remaining $m$ people. – JMoravitz May 07 '16 at 16:13
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    Possible duplicate of Vandermonde's Identity : Summations with binomial coefficients – JMoravitz May 07 '16 at 16:14
  • Also note: your summation should have gone from $j=0$ to $k$, not all the way up to $n$, though effectively this makes no difference because when $j>k$ the $\binom{m}{k-j}$ term becomes zero. – JMoravitz May 07 '16 at 16:16

3 Answers3

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It's coefficient of $x^k$ in $(1+x)^{m+n}$ or see Vandermondes identity as I see you don't need binomial proof .

Archis Welankar
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I was able to prove the result without using Mathematical Induction. Here are my workings:

Consider: $$ \begin{equation*} (1+x)^n(1+x)^m = \Bigg(1+\binom{n}{1}x + \binom{n}{2}x^2+...+\binom{n}{n}x^n\Bigg)\Bigg(1+\binom{m}{1}x + \binom{m}{2}x^2+...+\binom{m}{m}x^n\Bigg) \end{equation*} \tag{1} $$ Now, coefficient of $x^k$ from $(1)$ is: $$ \binom{n}{0}\binom{m}{k} + \binom{n}{1}\binom{m}{k-1}+...\binom{n}{k}\binom{m}{0} \tag{A} $$ But, $$ (1+x)^n(1+x)^m=(1+x)^{m+n} \tag{2} $$ Again, coefficient of $x^k$ from $(2)$ is; $$ \binom{n+m}{k} \tag{B} $$ Finally, from $(A)$ and $(B)$, we get: \begin{align*} &\binom{n}{0}\binom{m}{k} + \binom{n}{1}\binom{m}{k-1}+...\binom{n}{k}\binom{m}{0} = \binom{n+m}{k}\\ &\implies \sum_{j=0}^{n}\binom{n}{j}\binom{m}{k-j} = \binom{m+n}{j} \end{align*}

Kaustav Sen
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Using the recurrence relation for Pascal's Triangle, the inductive step can be shown as $$ \begin{align} \sum_{j=0}^n\binom{n}{j}\binom{m}{k-j} &=\sum_{j=0}^n\binom{n}{j}\left[\binom{m-1}{k-j}+\binom{m-1}{k-j-1}\right]\tag{1}\\ &=\sum_{j=0}^n\binom{n}{j}\binom{m-1}{k-j}+\sum_{j=0}^n\binom{n}{j}\binom{m-1}{k-j-1}\tag{2}\\ &=\binom{n+m-1}{k}+\binom{n+m-1}{k-1}\tag{3}\\ &=\binom{n+m}{k}\tag{4} \end{align} $$ Explanation:
$(1)$: use Pascal's Triangle recurrence
$(2)$: separate sums
$(3)$: use inductive hypothesis for case $n+m-1$
$(4)$: use Pascal's Triangle recurrence

robjohn
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