# Combinatorial Identities : Possible Simplification?

Hi everybody,

This is my first question so I hope I will correctly be following the rules!

I am looking for a simplification of the expression

$$m! \sum_{k=0}^n \binom{n}{k} \binom{\alpha k}{m} x^k,$$ where $n,m$ are integers and $0 < \alpha < 1$. Is it a known generating function? Or does it exist a simpler expression?

In the same order I am also trying to simplify the expression

$$\sum_{k=1}^n k! \binom{\alpha l}{k} S(n,k) x^k,$$

where $l$ is an integer?

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General mathematics doesn't seem like an appropriate tag. – Jim Conant Jul 25 '11 at 15:24
I am sorry, this is my first post! But I didn't know which tag I should use... – David Jul 26 '11 at 9:12

Not sure that it helps, but letting $$A_{m,n}(x) = m! \sum_{k=0}^n \binom{n}{k} \binom{\alpha k}{m} x^k$$ be the quantity you're studying, the generating function of $A_{m,n}(x)$ with respect to $m$ comes out to (if I haven't made an algebra error) $$\sum_{m=0}^\infty A_{m,n}(x)\frac{y^m}{m!} = (1+x(1+y)^\alpha)^n.$$

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Thanks for the answer! I can't see how you have this result though... However, I kind need a closed expression for $A_{m,n}(x)$ because I would need to take an expectation on $n$ (which is random)! Would you know where I could possibly find the answer? I have been looking for identities involving the generalization of falling factorials but I couldn't find any litterature... – David Jul 26 '11 at 9:10
The formula is trivial, just reverse the order of summation and use the binomial theorem twice. What makes you think that your series will have a simple closed expression. Most series don't. – Joe Silverman Jul 26 '11 at 12:36