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, 2022

Combinatorics Difficulty 6.1 National Olympiad Prove it Bulgaria

We will refer to a permutation σ\sigma of the numbers from 1 to 10 as bad, if there exist three indices i,j,ki, j, k, such that 1i<j<k101 \le i < j < k \le 10, but σ(j)<σ(k)<σ(i)\sigma(j) < \sigma(k) < \sigma(i), and as good, otherwise. Determine the number of the good permutations.

Solution

We will derive a closed formula for the good permutations of the numbers from 1 to nn. Consider a good permutation σ\sigma, where j=σ1(1)j = \sigma^{-1}(1) is an arbitrary index inbetween 1 and nn. Then, the first j1j-1 positions of σ\sigma contain the numbers 2,3,,j2, 3, \ldots, j, forming a good permutation of j1j-1 elements, while the last njn-j positions of σ\sigma contain the numbers j+1,,nj+1, \ldots, n, forming a good permutation of njn-j elements. It is an easy exercise to check that the above condition is not only necessary but also sufficient for the construction of an arbitrary good permutation of nn elements. Hence, denoting by SnS_n the corresponding number of good permutations, we derive that (Sn)n0(S_n)_{n \ge 0} satisfies S0=1S_0 = 1 and Sn=j=0n1SjSn1jS_n = \sum_{j=0}^{n-1} S_j S_{n-1-j}, n1\forall n \ge 1. Therefore, Sn=1n+1(2nn)S_n = \frac{1}{n+1} \binom{2n}{n} is the nn-th Catalan number. In particular,

S10=111(2010)=1213202310=4131719=16796. S_{10} = \frac{1}{11} \binom{20}{10} = \frac{12 \cdot 13 \cdots 20}{2 \cdot 3 \cdots 10} = 4 \cdot 13 \cdot 17 \cdot 19 = 16796.

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