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How to prove $${n \choose m}-{n \choose m+1} +{n \choose m+2}-\dots+{n \choose n}= {n-1 \choose m-1}$$

I tried proving this using binomial expansion adding and subtracting the first (m-1) terms but could not get the result. Is there another way to go about this?

RobPratt
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stat1809
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    This is basically a duplicate of https://math.stackexchange.com/questions/887960/truncated-alternating-binomial-sum$. The only difference is that the linked question about $\binom n0-\binom n1+\binom n2-\dots\pm \binom nm$, summing from $0$ to $m$, whereas you are summing from $m$ to $n$. But this makes no difference, because of the symmetry $\binom nk=\binom n{n-k}$. – Mike Earnest Jul 21 '24 at 18:55
  • Though in this case, $n-m$ has to be even so the $\binom nn$ term has a $+$ sign. – peterwhy Jul 21 '24 at 21:54

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