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I asked a question here which was closed for being a duplicate, and though I have edited the question to clarify why it was not it has not been reopened. To restate: the question I am asking is a proof by induction for ${m+n \choose m+r} = \sum\limits_{k=0}^{m}{m \choose k}\left({n \choose r+k}\right)$, not for $\binom{m+n}r = \sum_{k=0}^r \binom mk \binom n{r-k}$. If the second identity can be used to prove the first it is not clear to me how.

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    $\begingroup$ Sometimes mistakes are made in overlooking subtle (or not so subtle) differences in proposed duplicate Questions, and along with other such disputes there is a canonical thread where Requests for Reopening can be posted. As a first step (one you've already taken apparently), the Question's body is edited to make the distinction between target and proposed duplicate clearer. So I'll suggest using that canonical thread (posting an "answer" there) as the next step. $\endgroup$ – hardmath Oct 10 at 4:42
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    $\begingroup$ It does follow from that identity simply by using the reflection of Pascal's triangle. In other words: $$\binom{m+n}{m+r}=\binom{m+n}{n-r},\quad\binom{n}{r+k}=\binom{n}{n-r-k},$$ and you're done. The range of summation is a non-problem because you only need to include the non-zero terms anyway. $\endgroup$ – Jyrki Lahtonen Oct 10 at 5:07
  • $\begingroup$ For those who don't know the "reflection of Pascal's triangle", the mentioned post is not a duplicate of the linked one. There should be, if any, better "duplicate" target. $\endgroup$ – Jack Oct 10 at 20:53
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Let $m=2$, $r=3$, $n=4$; then the second equation says $${6\choose3}=\sum_0^3{2\choose k}{4\choose3-k}={2\choose0}{4\choose3}+{2\choose1}{4\choose2}+{2\choose2}{4\choose1}$$ Now let $m=2$, $r=1$, $n=4$; then the first equation says $${6\choose3}=\sum_0^2{2\choose k}{4\choose k+1}={2\choose0}{4\choose1}+{2\choose1}{4\choose2}+{2\choose2}{4\choose3}$$ These are obviously the same, since ${4\choose3-k}={4\choose k+1}$. I expect you can find a transformation that works in general to turn the one sum into the other.

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    $\begingroup$ With all due respect, this is why I cast my vote to reopen. Even if the problems are identical, it requires a solutions' worth of text to make that point obvious to the casual reader. $\endgroup$ – Matthew Daly Oct 10 at 18:50

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