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Harvard-MIT Mathematics Tournament

Combinatorics Difficulty 5.4 AIME, harder Prove it United States

Problem:
Let w=w1,w2,…,w6w = w_{1}, w_{2}, \ldots, w_{6} be a permutation of the integers {1,2,…,6}\{1, 2, \ldots, 6\}. If there do not exist indices i<j<ki < j < k such that wi<wj<wkw_{i} < w_{j} < w_{k} or indices i<j<k<li < j < k < l such that wi>wj>wk>wlw_{i} > w_{j} > w_{k} > w_{l}, then ww is said to be exquisite. Find the number of exquisite permutations.

Solution

Solution:
Given a permutation w=w1,…,wnw = w_{1}, \ldots, w_{n} for some nn, call a sequence wi1,wi2,…,wimw_{i_{1}}, w_{i_{2}}, \ldots, w_{i_{m}} an increasing subsequence if i1<⋯<imi_{1} < \cdots < i_{m} and wi1<⋯<wimw_{i_{1}} < \cdots < w_{i_{m}}. Define decreasing subsequences similarly. Let is(w)is(w) denote the length of the longest increasing sequence and ds(w)ds(w) denote the length of the longest decreasing sequence. We wish to find the number of permutations for n=6n = 6 such that is(w)≤2is(w) \leq 2 and ds(w)≤3ds(w) \leq 3. We note here that 6=2×36 = 2 \times 3 is not a coincidence.

The main idea of his proof is as follows. Consider a p×qp \times q rectangle. A Young Tableau is an assignment of 1,2,…,pq1, 2, \ldots, pq, one to each unit square of the rectangle, such that every row and column is in increasing order. There is a bijection between set of exquisite permutations and pairs of Young Tableaux. Since the number of ways to write 1,2,…,61, 2, \ldots, 6 on a 2×32 \times 3 rectangle with every row and column in increasing order is 55, there are exactly 2525 exquisite permutations.

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