Find the number of permutations π of {1,2,...,15} such that π ( j + 1 ) = π ( j ) + 1 for j = 1 , 2 , . . . , 1 4 .
Try other interesting combinatorics in my set Hard
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+1 for the solution and the problem! A problem I posted is very similar to this one, here it is - Vibgyor . I myself used the mobius inversion formula (it reduces to the principle of inclusion-exclusion here), it is a great trick to have up your sleeve!
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Thanks! I actually just solved Vibgyor using this same method too :)
The inclusion-exclusion form given above is a result of the mobius inversion (for sets) of the class of restricted permutations. I used it in an earlier analysis as well.
Really,you did great in using generating functions.One may also use inclusion-exclusion to prove a generalization-
The number of permutations π of 1 , 2 , . . , n such that π ( j + 1 ) = π ( j ) + 1 for j = 1 , 2 . . , n − 1 is D n + D n − 1 where D n is the n -th Derangement number.
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This is a tough problem. We can view this as a problem of generating all permutations, removing everything that has two consecutively placed elements, then removing everything that has three consecutively placed elements, adding back in the overlap between two and three consecutively placed elements permutations, ad infinitum.
This is inclusion exclusion, and we shall encode a consecutive run as a bunch of ∘ s. For instance, 1 2 3 5 4 6 8 7 ≡ 1 ∘ ∘ 5 4 6 8 7 . Let δ ( π ) be the number of ∘ in π , then we want our generating function to satisfy f ( z ) = π ∑ ( − 1 ) δ ( π ) z ∣ π ∣
This can be done by setting ∘ = − 1 in the construction f ( z , ∘ ) = P ( 1 − z ∘ z ) where P ( z ) = ∑ k k ! z k counts the permutations. The desired solution is [ z n ] f ( z , 1 ) = [ z n ] k ∑ k ! ( 1 + z z ) k
This OGF expands to z + z 2 + 3 z 3 + 1 1 z 4 + 5 3 z 5 + 3 0 9 z 6 + 2 1 1 9 z 7 + 1 6 6 8 7 z 8 + 1 4 8 3 2 9 z 9 + 1 4 6 8 4 5 7 z 1 0 + 1 6 0 1 9 5 3 1 z 1 1 + 1 9 0 8 9 9 4 1 1 z 1 2 + 2 4 6 7 0 0 7 7 7 3 z 1 3 + 3 4 3 6 1 8 9 3 9 8 1 z 1 4 + 5 1 3 1 3 7 6 1 6 7 8 3 z 1 5 + O ( z 1 6 ) so the solution is 5 1 3 1 3 7 6 1 6 7 8 3 .