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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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    $\begingroup$ 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... $\endgroup$
    – darij grinberg
    Commented Apr 29, 2013 at 14:53
  • $\begingroup$ 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? $\endgroup$
    – Barry Cipra
    Commented Apr 29, 2013 at 15:00
  • $\begingroup$ @Barry Cipra you're right. It's the players who are reordered, not the choices. $\endgroup$
    – Oren
    Commented Apr 29, 2013 at 15:03
  • $\begingroup$ Following up on darij grinberg's comment leads to oeis.org/A008279 $\endgroup$
    – Barry Cipra
    Commented Apr 29, 2013 at 15:10
  • $\begingroup$ I still don't get it. $\endgroup$
    – Oren
    Commented Apr 29, 2013 at 15:43

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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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  • $\begingroup$ I guess you meant $e_p$ in the second formula. $\endgroup$
    – Oren
    Commented Apr 29, 2013 at 16:28
  • $\begingroup$ so a good upper bound would be: $p!x^pe^\frac{1}{x}$ $\endgroup$
    – Oren
    Commented Apr 29, 2013 at 16:39

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