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HMMT February

Combinatorics Difficulty 5.2 AIME, harder Find the answer

A function f:A→Af: A \rightarrow A is called idempotent if f(f(x))=f(x)f(f(x))=f(x) for all x∈Ax \in A. Let InI_{n} be the number of idempotent functions from {1,2,…,n}\{1,2, \ldots, n\} to itself. Compute ∑n=1∞Inn!\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)kn−kA_{k, n}=\binom{n}{k} k^{n-k} for 1≤k≤n1 \leq k \leq n because there are (nk)\binom{n}{k} ways to choose the fixed points and all n−kn-k remaining elements must map to fixed points, which can happen in kn−kk^{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=1∞∑k=1nkn−kk!(n−k)!=∑k=1∞1k!∑n=k∞kn−k(n−k)!=∑k=1∞1k!∑n=0∞knn!=∑k=1∞1k!ek=ee−1=\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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