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$$ \binom{r}{r} + \binom{r+1}{r}+\binom{r+2}{r} + \cdots + \binom{n}{r}=\binom{n+1}{r+1}$$

I knew that I had to prove from RHS and LHS RHS simply like: take $r+1$ out of $n+1$ elements But How can I write about LHS?

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closed as off-topic by Eevee Trainer, Shailesh, Leucippus, darij grinberg, Parcly Taxel Feb 27 at 13:53

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  • $\begingroup$ use the combinatorics formula: $\frac{n!}{(n-r)! r!}$ $\endgroup$ – user29418 Feb 21 at 3:21
  • $\begingroup$ @user29418 can you give more details? $\endgroup$ – Math askers510520 Feb 21 at 3:27
  • $\begingroup$ $\frac{r!}{0!r!} + \frac{(r+1)!}{1!r!} + ...$, let $n=r, r=r$ for the first one; $n=r+1, r=r$ for the second one, etc. At that point it's all algebra. $\endgroup$ – user29418 Feb 21 at 3:38
  • $\begingroup$ For an algebraic argument, first expand out the binomials and cancel out factorials. The formula for $\binom nr = \frac{n!}{(n-r)!r!}$ is given above. $\endgroup$ – астон вілла олоф мэллбэрг Feb 21 at 3:50
  • $\begingroup$ See math.stackexchange.com/questions/1490794/… . $\endgroup$ – darij grinberg Feb 21 at 4:15
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In choosing $r+1$ of elements $1,2,\dots,n+1,$ suppose the the last element we choose is element $k.$ where $r+1\leq k\leq n+1.$ Then we must choose $r$ of the preceding $k-1$ elements. The total number of ways to do this is the sum on the left-hand side.

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