I am trying to find the number of permutations $P(10,10)$ and we have the condition that $\sigma(2i) = 2i$ for every $i\in [5]$. I though about inclusion exclusion or that we essentially hold 5 points fixed and permute the other 5 points so we would have $5!$ ways. Can anybody tell me if I am going down the right path?
2026-04-04 09:33:21.1775295201
Permutations with fixed point conditions
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Yes, you are right. If you fix the $5$ even, you just have to permute the $5$ odd in $5!$ ways. Inclusion-exclusion would be useful if you have at least one even fixed.