Binomial Theorem and Sum Conversion

Number Cruncher 20 · #1 $Old$ August 25th, 2008, 04:22 AM

Hi,

I have been asked to derive a simple closed form expression for the
following sum:

n
∑ (-1)^(m+i) *(nCi)*(iCm) = S n, m
i = m

[with m = 0,1,2,....,n ; n= 0,1,2,....] and C is the combinations symbol.

I know that we are meant to be using the binomial theorem but I dont quite know where to go from there. Any help would be much appreciated.

mr fantastic · #2 $Old$ August 25th, 2008, 04:44 AM

Quote:

Originally Posted by Number Cruncher 20 $View Post$

Hi,

I have been asked to derive a simple closed form expression for the
following sum:

n
∑ (-1)^(m+i) *(nCi)*(iCm) = S n, m
i = m

[with m = 0,1,2,....,n ; n= 0,1,2,....] and C is the combinations symbol.

I know that we are meant to be using the binomial theorem but I dont quite know where to go from there. Any help would be much appreciated.

Three hints:

1. Prove that $\sum_{i=0}^{n} (-1)^i {n \choose i}$ is equal to zero if $n \neq 0$ and is equal to 1 if $n = 0$ .

2. Prove that ${n \choose k} {k \choose l} = {n \choose l} {n - l \choose k - l}$ for $l \leq k$ .

3. Deduce from 1. and 2. that your expression is equal to zero if $m \neq n$ and is equal to 1 if $n = m$ .

Number Cruncher 20 · #3 $Old$ August 25th, 2008, 05:56 AM

Yes I see now, thanks a lot for the help!