Recall that a derangement of a set is a permutation of having no fixed points, i.e. . If the number of derangements of is , show that
Here is another proof.
First, I prove the statement in the hint that I gave. Consider the exponential generating function
of the sequence . By the basic properties of exponential generating functions, the product of the exponential generating functions of two sequences is the exponential generating function of their binomial convolution. So we have
and so . Extracting the coefficient of on both sides we obtain as required.
Now let us partition the set of permutations of as follows; let denote those permutations which have precisely fixed points, for . Clearly there are ways of picking the elements which are to be fixed, and the remaining elements have to be deranged. So . Since the are clearly distinct we have
because . Now just apply the binomial involution just proved!