Consider $P(n)$ as a number of $n$-permutations, which each cycle have even length, and $N(n)$ as a number of $n$-permutations, which each cycle have odd length. Calculate $P(2n)-N(2n)$
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Yes, we are considering P(2n) - N(2n) where n is a natural number, and we are considering all types of cycle – Steve Mar 21 '13 at 18:33
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Isn't this the same as was discussed here? – Marko Riedel Mar 21 '13 at 21:14
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Yes, it's the same question. – David Moews Mar 21 '13 at 22:02