The problem: given numbers *n* and *k*, find the number of functions such that for each *x* with 1 ≤ *x* ≤ *k* there is a number *r* with *f*^{[r]}(*x*) = 1 and for each *x* with *k* < *x* ≤ *n* there is **no** such number *r*.

First, observe that *f*(*x*) with *x* > *k* can be set to anything in {*k* + 1... *n*}, independently to what *f* is in {1... *k*}. This means that we can just calculate the number of such functions *f* restricted to {1, ... *k*} and multiply by (*n* - *k*)^{n - k} by the multiplication principle.

But how do we find how many functions there are that come back to 1 after sufficiently many iterations? If you draw some pictures, you may think about labeled trees: there are *k*^{}(*k* - 2) labeled trees on *n* vertices by [Caley's theorem](http://en.wikipedia.org/wiki/Tree_(graph_theory)#Labeled_trees). Each function corresponds to a unique graph with vertices 1... *n* and an edge *i*, *j* if *f*(*i*) = *j*. The problem statement says that you can come back to 1 if you follow the edges, no matter where you start, this means that there are no directed cycles except the ones containing 1.

There is a problem though. Caley's formula counts undirected trees, and we have something that is not even a DAG (because there can be cycles containing 1). We can do a trick to make the thing a DAG — just remove the edge — this breaks the cycle with 1 inside and makes the graph a DAG. Also, it turns out that there are no undirected cycles as well: if there would be an undirected circle, it cannot be a directed one, and this means that some vertex is the cicle has two edges going out from it which is a contradiction because *f* is a function.

This was a sketch of the proof that the graph is a (unique) labeled tree with an extra edge

*f*(1) can have *k* values, so the answer is: there are *k*·*k*^{(k - 2)} ways to choose where the first *k* values go and (*n* - *k*)^{(n - k)} (independent) ways to choose the last *n* - *k*. Thus the complete answer is *k*^{(k - 1)}·(*n* - *k*)^{(n - k)}.

There are some gaps in the proof: we didn't show that every tree together with the extra edge corresponds to a function, and that the correspondence is unique.

For 1 <= x <= k, I came up with this solution: out of k^k ways we must remove (k-1)*k^(k-1) ways because for each p[i] = i (1 < i <= k) we cannot go back to home 1 from position i. And k^k — (k-1)*k^(k-1) is k^(k-1).