Drexel28, This is a new problem derived from another thread.
So, enjoy this problem!
A Challenge Problem:
Let be all the permutations of {1,2,...,n}.
for any given permutation , if , we say is a fix point of .
A permutation may have several fix points, let be the set of fix points of the permutation .
Problem: How many permutations are there whose fix-points set has exactly k elements(points)? k=0,1,...,n.
Solution: Fix the elements of that you want to fix. Then, the number of permutations with those fixed points are precisely the amount of bijective maps one can form from points onto itself with . These are the derangements of . So, for the fixed elements there are permutations. Thus, the total number of permutations is the number of ways you can pick the elements. Thus, the answer is .
Remark: It can be proven that