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I'm looking for a simple identity for the formula:

$$ \sum_{n = 0}^{p} \binom{p}{n} \cdot n! \cdot x^n $$

In words, I have $p$ "players" who can choose to play or not (every player is represented by a unique id). Those who chose to play are lined up in all possible orders. Then every playing player picks an element out of $x$ possibilities (with repetition allowed), in addition to his id.

How many sequences can we get? Is there a simple solution for this series? If not, what is the closest upper limit you can think of?

I haven't touched combinatorics for a long time, so there could be a simple identity that I'm missing...

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There seems to be some capacity for finding "2-dimensional sequences" (sequences with 2 indices) on OEIS, though I have never used it. This one, though, reminds me too much of the truncated exponential series (more precisely, if you divide your sum by $p!x^p$ and replace $x$ by $1/x$, you get a truncated exponential series) to have a nice closed form... –  darij grinberg Apr 29 '13 at 14:53
    
I think a better way to say this in words is that you have $p$ people who can choose to play or not, those who opt to play are lined up in all possible orders, then each players picks (with repetitions allowed) an element out of $x$ possibilities. Or is there some reason you want to think of the players' choices as getting reordered? –  Barry Cipra Apr 29 '13 at 15:00
    
@Barry Cipra you're right. It's the players who are reordered, not the choices. –  Oren Apr 29 '13 at 15:03
    
Following up on darij grinberg's comment leads to oeis.org/A008279 –  Barry Cipra Apr 29 '13 at 15:10
    
I still don't get it. –  Oren Apr 29 '13 at 15:43

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up vote 0 down vote accepted

Darij's comment...

Tthe truncated exponential series: $$ e_p(z) = \sum_{n=0}^p\frac{z^n}{n!} $$ Your sum is $$ p!x^pe_p(1/x) $$

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I guess you meant $e_p$ in the second formula. –  Oren Apr 29 '13 at 16:28
    
so a good upper bound would be: $p!x^pe^\frac{1}{x}$ –  Oren Apr 29 '13 at 16:39

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