# Binomial coefficients

• April 5th 2008, 08:33 AM
james_bond
Binomial coefficients
$
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=?
$
• April 5th 2008, 08:38 AM
Mathstud28
If you are asking
what that series is equal to i think its $\sum_{n=0}^{k}{k\choose n}(k-n)$
• April 5th 2008, 08:47 AM
TrevorP
This what I got in terms of a summation.
$
k^k + \sum_{n=1}^{k}{k\choose n}(k-n)^k(-1)^n
$
• April 5th 2008, 08:48 AM
Mathstud28
Yeah
hes right I did not see the alternating signs
• April 5th 2008, 12:11 PM
james_bond
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 ( $\sum_{n=0}^{k}{k\choose n}(k-n)^k(-1)^n$)...
• April 5th 2008, 12:21 PM
Plato
Quote:

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.