Maths Olympiad Prep

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HMMT February

Combinatorics Difficulty 5.3 AIME, harder Find the answer

Let S7S_{7} denote all the permutations of 1,2,…,71,2, \ldots, 7. For any π∈S7\pi \in S_{7},let, let f(π)f(\pi) be the smallest positive integer isuchthatπ(1),π(2),…,π(i) such that \pi(1), \pi(2), \ldots, \pi(i) is a permutation of 1,2,…,i1,2, \ldots, i. Compute ∑π∈S7f(π)$.\sum_{\pi \in S_{7}} f(\pi)\$.

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

Solution

Extend the definition of ff to apply for any permutation of 1,2,…,n1,2, \ldots, n, for any positive integer nn. For positive integer nn, let g(n)g(n) denote the number of permutations π\piof of 1,2, …,\ldots, nsuchthat such that f(π)=nf(\pi)=n.Wehave. We have g(1)=1.Forfixed. For fixed n, k(with (with k ≤\leq n),thenumberofpermutationsπ ), the number of permutations \pi of 1,2,…,n1,2, \ldots, n such that f(π)=kf(\pi)=k is g(k)(n−k)g(k)(n-k) !. This gives us the recursive formula g(n)=g(n)= n!−∑k=1n−1g(k)(n−k)n!-\sum_{k=1}^{n-1} g(k)(n-k) !. Using this formula, we find that the first 7 values of gg are 1,1,3,13,71,461,34471,1,3,13,71,461,3447. Our sum is then equal to ∑k=17k⋅\sum_{k=1}^{7} k \cdot g(k)(7-k) !. Using our computed values of g$, we get that the sum evaluates to 29093 .

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