CombinatoricsDifficulty 5.2AIME, harderFind the answer
A function f:A→A is called idempotent if f(f(x))=f(x) for all x∈A. Let In be the number of idempotent functions from {1,2,…,n} to itself. Compute ∑n=1∞n!In.
A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.
Solution
Let Ak,n denote the number of idempotent functions on a set of size n with k fixed points. We have the formula Ak,n=(kn)kn−k for 1≤k≤n because there are (kn) ways to choose the fixed points and all n−k remaining elements must map to fixed points, which can happen in kn−k ways. Hence n=1 I n n! = n=1 k=1 n A k, n}}{n!} =∑n=1∞∑k=1nk!(n−k)!kn−k=∑k=1∞k!1∑n=k∞(n−k)!kn−k=∑k=1∞k!1∑n=0∞n!kn=∑k=1∞k!1ek=ee−1
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