For example, there are kn total ways of assigning k numbers to n numbers, but not all of the assignments use all k numbers. Let f(n,k) be the number of ways of assigning the numbers so that all k numbers are used. Then $\sum\limits_{i = 1}^k {C(k,i)f(n,i)}$ = kn.
There is a recurrence relationship that can be used to compute f(n,k).
f(n,k) = k(f(n-1,k) + f(n-1, k-1)), where f(n,0)=f(0,k)=0 and f(1,1)=1.
Also, for what it's worth, f(n,n) = n!.
With all this going for it, there must at least be a way of approximating f(n,k).
