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PrintSAUDI ARABIAN MATHEMATICAL COMPETITIONS
Saudi Arabia counting and probability
Problem
Find the number of permutations of the first positive integers satisfying the following two conditions: 1. for all . 2. There are exactly two indices with such that and .
Solution
For each positive integer , we denote as the number of permutations of the first positive integers that satisfy the condition We call these permutations "nice".
First, we shall prove that for all . Let be the index such that . We have which implies that , and so on, we have . All numbers after are forming a nice permutation, so in case , the number of nice permutations is . Note that if , we have only nice permutation: . Therefore, we get the following formula: It is easy to compute , , then by induction, we have for all .
Back to the original problem, we suppose that and . Thus, and so on, then we have . But in fact, which implies that the equality must occur. So for all . But there are only indices like that so we have .
Continue, , but , so . Then, By the same way, it is clear to check that . Similarly, we also have . Therefore, the lengths of the two sequences and are the same. Hence, So the given nice permutation looks like Clearly, the subsequence forms a nice permutation of length and the same with the subsequence Therefore, the number of permutations satisfying the given condition is
First, we shall prove that for all . Let be the index such that . We have which implies that , and so on, we have . All numbers after are forming a nice permutation, so in case , the number of nice permutations is . Note that if , we have only nice permutation: . Therefore, we get the following formula: It is easy to compute , , then by induction, we have for all .
Back to the original problem, we suppose that and . Thus, and so on, then we have . But in fact, which implies that the equality must occur. So for all . But there are only indices like that so we have .
Continue, , but , so . Then, By the same way, it is clear to check that . Similarly, we also have . Therefore, the lengths of the two sequences and are the same. Hence, So the given nice permutation looks like Clearly, the subsequence forms a nice permutation of length and the same with the subsequence Therefore, the number of permutations satisfying the given condition is
Final answer
2^{2012}
Techniques
Recursion, bijectionInduction / smoothing