Given $k \leq n$ positive integers, how to simplify ${ n \choose k } + { n-1 \choose k-1 } + \dots + { n-k \choose 0 }$?
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5${5 \choose 2}+{4 \choose 1}+{3 \choose 0} = 10+4+1 = 15$ rather than $17$. Look at Pascal's triangle to recognise your results. $ { n \choose k } + { n-1 \choose k-1 } + \dots + { n-k \choose 0 } = { n+1 \choose k }$ – Henry Oct 11 '20 at 20:06
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this is probably provable by induction – person Oct 11 '20 at 20:16
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Indeed very simple. Thank you. – Aladin Oct 11 '20 at 20:19
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1The induction step is ${ n+1 \choose k+1 } +{ n \choose k } + { n-1 \choose k-1 } + \dots + { n-k \choose 0 } = { n+1 \choose k+1 }+ { n+1 \choose k } = { n +2\choose k +1}$ – Henry Oct 11 '20 at 20:21
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"This seems to have binomial-coefficient-like behavior, but I can't figure the exact form. " That made my day. – Servaes Oct 11 '20 at 23:30
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Does this answer your question? Proof of the Hockey-Stick Identity: $\sum\limits_{t=0}^n \binom tk = \binom{n+1}{k+1}$ – zhoraster Oct 12 '20 at 06:18
2 Answers
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$$\binom{n+1}{k}-\binom{n}{k}=\binom{n}{k-1}$$ $$\binom{n}{k-1}-\binom{n-1}{k-1}=\binom{n-1}{k-2}$$ $$\binom{n-1}{k-2}-\binom{n-2}{k-2}=\binom{n-2}{k-3}$$ $$\vdots$$ $$\binom{n-k+2}{1}-\binom{n-k+1}{1}=\binom{n-k+1}{0}$$ $$\binom{n-k+1}{0}-\binom{n-k}{0}=1-1=0$$
Summing we get $$\binom{n+1}{k}-\left(\binom{n}{k}+\binom{n-1}{k-1}+\binom{n-2}{k-2}+\cdots+\binom{n-k}{0}\right)=0$$
Hence simpler form for your expression is $\binom{n+1}{k}$.
QED
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Applying symmetry twice and the hockey-stick identity, we have $$\sum_{j=0}^k \binom{n-k+j}{j} = \sum_{j=0}^k \binom{n-k+j}{n-k} = \binom{n+1}{n-k+1} = \binom{n+1}{k}$$
RobPratt
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