CombinatoricsDifficulty 5.9AIME, harderProve itUnited States
Problem:
Let [n] denote the set of integers {1,2,…,n}. We randomly choose a function f:[n]→[n], out of the nn possible functions. We also choose an integer a uniformly at random from [n]. Find the probability that there exist positive integers b,c≥1 such that fb(1)=a and fc(a)=1. (fk(x) denotes the result of applying f to xk times).
Solution
Solution:
Answer: n1
Given a function f, define N(f) to be the number of numbers that are in the same cycle as 1 (including 1 itself), if there is one, and zero if there is no such cycle. The problem is equivalent to finding E(N(f))/n. Note that P(N(f)=k)=nn−1⋅nn−2⋯nn−k+1⋅n1 and it suffices to compute ∑k=1nPk where Pk=nkP(N(f)=k). Observe that PnPn−1⇒Pn+Pn−1Pn−2⇒Pn+Pn−1+Pn−2⇒∑k=1nPk=(nn−1⋅nn−2⋯n3⋅n2⋅n1)⋅nn=(nn−1⋅nn−2⋯n3⋅n2)⋅nn−1=(nn−1⋅nn−2⋯n3⋅n2)=(nn−1⋅nn−2⋯n3)⋅nn−2=(nn−1⋅nn−2⋯n3)⋯=1⋅n1⋅n1⋅n1⋅n1⋅n1 Therefore the answer is n1.
Want a route through all this instead of an archive? The track
puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.
Source: MathNet,
licensed CC-BY-4.0.
Statement reproduced verbatim; metadata (topic, difficulty) added by this project.