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NMO Selection Tests for BMO and IMO

Combinatorics Difficulty 7.8 National Olympiad, round 2 Prove it Romania

Let nn be a positive integer, let SnS_n be the set of all permutations of the set {1,2,…,n}\{1, 2, \dots, n\}, and, for each σ\sigma in SnS_n, let I(σ)={i:σ(i)≤i}I(\sigma) = \{i: \sigma(i) \le i\}. Evaluate the sum
∑σ∈Sn1∣I(σ)∣∑i∈I(σ)(i+σ(i)). \sum_{\sigma \in S_n} \frac{1}{|I(\sigma)|} \sum_{i \in I(\sigma)} (i + \sigma(i)).

Solution

Consider the involution of SnS_n which sends a permutation σ\sigma to the permutation σ∗\sigma^* defined by σ∗(i)=j\sigma^*(i) = j if and only if σ(n−j+1)=n−i+1\sigma(n-j+1) = n-i+1. Notice that, for each σ\sigma in SnS_n, the assignment i↦n−σ(i)+1i \mapsto n-\sigma(i)+1 defines a bijection from I(σ)I(\sigma) to I(σ∗)I(\sigma^*) whose inverse sends jj to n−σ∗(j)+1n-\sigma^*(j)+1. Consequently,
∑σ∈Sn1∣I(σ)∣∑i∈I(σ)(i+σ(i))==12∑σ∈Sn(1∣I(σ)∣∑i∈I(σ)(i+σ(i))+1∣I(σ∗)∣∑i∈I(σ∗)(i+σ∗(i)))=12∑σ∈Sn1∣I(σ)∣∑i∈I(σ)(i+σ(i)+(n−σ(i)+1)+(n−i+1))=(n+1)!. \begin{align*} \sum_{\sigma \in S_n} \frac{1}{|I(\sigma)|} \sum_{i \in I(\sigma)} (i + \sigma(i)) &= \\ &= \frac{1}{2} \sum_{\sigma \in S_n} \left( \frac{1}{|I(\sigma)|} \sum_{i \in I(\sigma)} (i + \sigma(i)) + \frac{1}{|I(\sigma^*)|} \sum_{i \in I(\sigma^*)} (i + \sigma^*(i)) \right) \\ &= \frac{1}{2} \sum_{\sigma \in S_n} \frac{1}{|I(\sigma)|} \sum_{i \in I(\sigma)} (i + \sigma(i) + (n - \sigma(i) + 1) + (n - i + 1)) \\ &= (n + 1)! . \end{align*}

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