Problem 45193. Fun with permutations

Created by William

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{"relationships":[{"relationshipType":"http://schemas.mathworks.com/matlab/code/2013/relationships/document","relationshipId":"rId1","target":"/matlab/document.xml"},{"relationshipType":"http://schemas.mathworks.com/matlab/code/2013/relationships/output","relationshipId":"rId2","target":"/matlab/output.xml"}],"parts":[{"partUri":"/matlab/document.xml","relationship":[],"contentType":"application/vnd.mathworks.matlab.code.document+xml","content":"<?xml version=\"1.0\" encoding=\"UTF-8\"?><w:document xmlns:w=\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\"><w:body><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>There are factorial(N) permutations of the numbers from 1 to N. For each of these permutations, we can find the set of indexes j that will return the numbers to their original order, 1:N. For example, if we generate a random permutation p with p = randperm(1:N) then the set of indices j that will return p to the original order p(j) = 1:N may be found by [~,j] = sort(p).</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/><w:jc w:val=\"left\"/></w:pPr><w:r><w:t>It sometimes happens that this set of indices j and the corresponding permutation p are the same. For example, if p = 1:N, then j = 1:N as well; and if p = N:-1:1, then j is also N:-1:1. However, for N = 3, there are 6 permutations in all, and it happens that only 4 of them have this property.</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/><w:jc w:val=\"left\"/></w:pPr><w:r><w:t>Your task is to determine, given the value of N, the number of permutations p of the array 1:N for which the permutation p and the indexes j are the same.</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/><w:jc w:val=\"left\"/></w:pPr><w:r><w:t>input: N, the range of the integers, output: The number of permutations of 1:N for which isequal(p,j)</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/><w:jc w:val=\"left\"/></w:pPr><w:r><w:t>HINT: For N &lt; 12, you can probably solve this by checking each of the individual permutations, but for larger N this may prove to be too time-consuming.</w:t></w:r></w:p></w:body></w:document>"},{"partUri":"/matlab/output.xml","contentType":"text/xml","content":"<?xml version=\"1.0\" encoding=\"UTF-8\" standalone=\"no\" ?><embeddedOutputs><metaData><evaluationState>manual</evaluationState><layoutState>code</layoutState><outputStatus>ready</outputStatus></metaData><outputArray type=\"array\"/><regionArray type=\"array\"/></embeddedOutputs>"}]}

There are factorial(N) permutations of the numbers from 1 to N. For each of these permutations, we can find the set of indexes j that will return the numbers to their original order, 1:N. For example, if we generate a random permutation p with p = randperm(1:N) then the set of indices j that will return p to the original order
p(j) = 1:N may be found by [~,j] = sort(p).It sometimes happens that this set of indices j and the corresponding permutation p are the same. For example, if p = 1:N, then j = 1:N as well; and if p = N:-1:1, then
j is also N:-1:1. However, for N = 3, there are 6 permutations in all, and it happens that only 4 of them have this property.Your task is to determine, given the value of N, the number of permutations p of the array 1:N for which the permutation p and the indexes j are the same.input: N, the range of the integers,
output: The number of permutations of 1:N for which isequal(p,j)HINT: For N &lt; 12, you can probably solve this by checking each of the individual permutations, but for larger N this may prove to be too time-consuming.

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Last Solution submitted on Jun 10, 2020

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Problem 45193. Fun with permutations

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