Binomial coefficients

**james_bond** · April 5th 2008, 07:33 AM

$ k^k - {k\choose 1}\left( {k - 1} \right)^k + {k\choose 2}\left( {k - 2} \right)^k - {k\choose 3}\left( {k - 3} \right)^k + \ldots + \left( { - 1} \right)^k {k\choose k}\left( {k - k} \right)^k=? $

**Mathstud28** · April 5th 2008, 07:38 AM

what that series is equal to i think its $\sum_{n=0}^{k}{k\choose n}(k-n)$

**TrevorP** · April 5th 2008, 07:47 AM

This what I got in terms of a summation.
$ k^k + \sum_{n=1}^{k}{k\choose n}(k-n)^k(-1)^n $

**Mathstud28** · April 5th 2008, 07:48 AM

hes right I did not see the alternating signs

**james_bond** · April 5th 2008, 11:11 AM

I suppose that the answer should be given in a closed form. I can see that it can be written shorter using the sum symbol ( $\sum_{n=0}^{k}{k\choose n}(k-n)^k(-1)^n$ )...

**Plato** · April 5th 2008, 11:21 AM

Originally Posted by james_bond

$ k^k - {k\choose 1}\left( {k - 1} \right)^k + {k\choose 2}\left( {k - 2} \right)^k - {k\choose 3}\left( {k - 3} \right)^k + \ldots + \left( { - 1} \right)^k {k\choose k}\left( {k - k} \right)^k=?$

That happens to count the number of surjections from a set of k to a set of k objects.
Or another way it counts surjections from a finite set of k objects onto itself. Those happen to be permutations.
How ways can we permute k objects. That is your answer.

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