1. ## 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=?
$

2. ## If you are asking

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

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

4. ## Yeah

hes right I did not see the alternating signs

5. 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 ( $\sum_{n=0}^{k}{k\choose n}(k-n)^k(-1)^n$)...

6. 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.