Is there a way to simplify : $\sum_{i=1}^{n} {n \choose i} {m \choose i}$?
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If $m=n$, we get $\binom{2n}n$ – Simply Beautiful Art Mar 18 '17 at 12:39
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What's the value of $m$? – Icycarus Mar 18 '17 at 12:40
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1@Icycarus what do you mean? m and n are random integers. – Nathan Mar 18 '17 at 12:42
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@SimplyBeautifulArt whats about case $m \neq n$? – Nathan Mar 18 '17 at 12:44
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@Nathan Beats me. I haven't seen this particular problem before, and it doesn't look very interesting to solve IMO. – Simply Beautiful Art Mar 18 '17 at 12:46
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2@SimplyBeautifulArt You have seen it before, it is the Vandermonde convolution identity with the first term missing.$\sum_{i=1}^{n} \binom{n,i} \binom{m,i} =\binom{n+m,n}-1$ & it is very interesting IMHO :-) – Donald Splutterwit Mar 18 '17 at 13:02
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@DonaldSplutterwit :-) Ah, I did not recognize. – Simply Beautiful Art Mar 18 '17 at 13:03
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@SimplyBeautifulArt It was the removal of the first term that got you ? – Donald Splutterwit Mar 18 '17 at 13:04
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@DonaldSplutterwit Nah, I just don't do combinatoric sums all the time. Its been a while for me. – Simply Beautiful Art Mar 18 '17 at 13:06
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Related : https://math.stackexchange.com/questions/855538/evaluate-sum-k-0n-n-choose-km-choose-k-for-a-given-n-and-m – Arnaud D. Nov 08 '18 at 11:15
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prove with induction that we get $$\sum_{i=1}^n\binom{n}{i}\binom{m}{i}={\frac { \left( m+1 \right) \left( n+m \right) !}{ \left( m+1 \right) !\,n!}}-1 $$
Dr. Sonnhard Graubner
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Thanks, right side can be simplified to:
$\frac{(n+m)!}{m!n!}-1 = {{n+m} \choose {m}} - 1$.
I am wondering how you come up with this right side?
– Nathan Mar 18 '17 at 12:58 -
i have this problem on my PC! i'm working with my students on combinatorics – Dr. Sonnhard Graubner Mar 18 '17 at 13:02
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Is there a way to came to the right side if a have no idea about how it looks? – Nathan Mar 18 '17 at 13:07
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Prof. Tutschke always say: try a number example to find a formula – Dr. Sonnhard Graubner Mar 18 '17 at 13:09
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When $n\leq m$ then $\sum_{k=0}^n \binom nk \binom m{k}$ counts all the ways to swap items between piles of $n$ and $m$ items; by doing counting selections of items in each pile to swap for each swap size ($k\in \{0..n\}$). Also when $n>m$ it counts the same by the convention that $\binom {m}{r}=0$ for all $r>m$.
We could do this by gathering all the items into one pile of $n+m$ and count ways to sort them back into piles of $n$ and $m$ items. This is $\binom {n+m}m$
Eg $\binom{5}2=10$
Thus as the count of ways to perform the same task must be the same:
$$\sum_{k=0}^n \binom n k\binom m k~=~ \binom {n+m}n ~=~ \binom {m+n}m ~=~ \sum_{k=0}^m \binom m k\binom n k$$
$~\\~\\~\\~\\~$
NB: Your series is indexed $1$, so does not include the term $\binom n0\binom m0$ which is 1.
$$\sum_{k=1}^n \binom n k\binom m k~=~ \binom {n+m}n-1 ~=~ \binom {m+n}m -1~=~ \sum_{k=1}^m \binom m k\binom n k$$
Graham Kemp
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