Maths Olympiad Prep

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Combinatorics Difficulty 5.2 AIME, harder Find the answer

A function f:AAf: A \rightarrow A is called idempotent if f(f(x))=f(x)f(f(x))=f(x) for all xAx \in A. Let InI_{n} be the number of idempotent functions from {1,2,,n}\{1,2, \ldots, n\} to itself. Compute n=1Inn!\sum_{n=1}^{\infty} \frac{I_{n}}{n!}.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Let Ak,nA_{k, n} denote the number of idempotent functions on a set of size nn with kk fixed points. We have the formula Ak,n=(nk)knkA_{k, n}=\binom{n}{k} k^{n-k} for 1kn1 \leq k \leq n because there are (nk)\binom{n}{k} ways to choose the fixed points and all nkn-k remaining elements must map to fixed points, which can happen in knkk^{n-k} ways. Hence n=1 I n n! = n=1 k=1 n A k,\text{n=1 I n n! = n=1 k=1 n A k,} n}}{n!} =n=1k=1nknkk!(nk)!=k=11k!n=kknk(nk)!=k=11k!n=0knn!=k=11k!ek=ee1=\sum_{n=1}^{\infty} \sum_{k=1}^{n} \frac{k^{n-k}}{k!(n-k)!} =\sum_{k=1}^{\infty} \frac{1}{k!} \sum_{n=k}^{\infty} \frac{k^{n-k}}{(n-k)!} =\sum_{k=1}^{\infty} \frac{1}{k!} \sum_{n=0}^{\infty} \frac{k^{n}}{n!} =\sum_{k=1}^{\infty} \frac{1}{k!} e^{k} =e^{e}-1

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