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$$\sum_{a+n-b=c}\binom{n}{a}\binom{n}{b}$$ What does this mean ? I found it here It is meant to sum up from where to where ?

Community
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reaffer
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  • There are implicit end values, since for example $\binom{n}{a}=0$ if $a\gt n$. – André Nicolas Nov 30 '15 at 21:12
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    Take all pairs $a,b$ of integers such that $a-b=c-n$. For only finitely many is $\binom{n}{a}\binom{n}{b}$ non-zero, so you can sum the value of the epxression even though there are technically infinitely many pairs $a,b$. – Thomas Andrews Nov 30 '15 at 21:13
  • @ThomasAndrews: I was about to write up a similar comment. But if $c = 0$, you can take any $n$ and then put $a = 0$ and $b = n$, giving infinitely many triples with $a + n - b = c$ and $\binom{n}{a}\binom{n}{b} = \binom{n}{0}\binom{n}{n} = 1 \cdot 1 = 1$. Presumably the inputs here must be $n$ and $c$ - this is not obvious from the context. – Rob Arthan Nov 30 '15 at 21:22
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    I'm assuming $c,n$ fixed. If they are also variables, then you are correct. (If $n$ is fixed, it is still a finite sum.) @RobArthan – Thomas Andrews Nov 30 '15 at 21:23
  • n is fixed. c goes from 0 to 2n. – fleablood Nov 30 '15 at 22:33
  • @fleablood: how do you know that when the OP hasn't chosen to share that information with us? – Rob Arthan Nov 30 '15 at 22:52
  • @RobArthan I looked at the page s/he linked to.. It was on that page. In fact it was part of a sum $\sum_{c=0}^{2n}(\sum_{a+n -b =c}{n \choose a}{n \choose b})x^c$ which is context would have made explaining what it meant a LOT easier. – fleablood Nov 30 '15 at 22:58

4 Answers4

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Assuming that $a,b, c$ are non-negative integers, $$\begin{align} \sum_{a+n-b=c}\binom{n}{a}\binom{n}{b}&=\sum_{a+n-b=c}\binom na\binom n{n-b}\\ &=\sum_{a^*}\binom na\binom n{c-a}\\ &=\binom {2n}c\qquad\qquad\text{by the Vandermonde Identity} \end{align}$$ where $a^*$ indicates the following ranges for $a$: $$\begin{cases} \begin{align} 0<c<=n: &\qquad a=0\quad\text{to}\quad c\\ n<c\leq 2n: &\qquad a=c-n\quad\text{to}\quad n \end{align}\end{cases}$$ such that $0\leq a,\;c-a \leq n$.

Hypergeometricx
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In context , both $a$ and $b$ are whole numbers less than or equal to $n$ $$\sum_{a+n-b=c}\binom{n}{a}\binom{n}{b}=\sum_{a=a_{min}}^{a_{max}} \binom n a \binom n{(n+a)-c}$$

where $a_{max}=\min\{ n, c-n \}$ and $a_{min}=\max\{ 0, c-n \}$

WW1
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You could write the sum over $a$ as $$\sum_{a= \max(0, c-n)}^{\min(n,c)}\binom{n}{a}\binom{n}{a+n-c}$$

Henry
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You sum it up on all possible values where (a + n - b=c) [In the context, n is a constant and c is a constant between 0 and 2n]. The conditions of a summation need not be a sequenced list. You can do $\sum_{p|36; p >0} p$ to mean $1 + 2 + 3 + 4 + 6 + 9 + 12 + 18 + 36$. This is in basically easier than writing $\sum_{i = 1}^{36} a_i$ where $a_i = i$ if $i|36$ and $a_i = 0$ if $i$ doesn't divide $36$.

So this is basically saying $\sum_{c = 0}^{2n}(\sum_{a = 0}^n \sum_{b = 0}^n{n \choose a}{n \choose b})$ *BUT ONLY IF b = n - c + a * or $\sum_{c=0}^{2n}(\sum_{a = \max(0, n-c)}^{\min(n,2n -c)} {n \choose a}{n \choose {n - c + a}})$

For example: if c = 0, this is a=0, b=n only. If c = v < n, this is a= 0...v; b=n,(n-1)....(n -v). If c = v > n, this is a = v-n...n; b = 0...(2n -v).

It's just easier to write as $\sum_{c=0}^{2n}(\sum_{a + n - b = c} {n \choose a}{n \choose b})$ and once you get used to it it is easier to understand.

Basically $\sum_{c=0}^{2n}(\sum_{a + n - b = c} {n \choose a}{n \choose b})$ = $\sum_{a=0}^0{n \choose a}{n \choose {a + n}} + \sum_{a=0}^1{n \choose a}{n \choose {a + n -1}} + \sum_{a=0}^2{n \choose a}{n \choose {a + n -2}} + .... \sum_{a=0}^n{n \choose a}{n \choose {a}} + \sum_{a=1}^n{n \choose a}{n \choose {a - 1}} + .... + \sum_{a=n-1}^n{n \choose a}{n \choose {a -n + 1}} +\sum_{a=n}^n{n \choose a}{n \choose {a -n }}$

Akiva Weinberger
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fleablood
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