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Max Alekseyev
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The answer is $$\binom nh d_{h,N},$$ where $d_{h,N}$ is the number of derangements of size $h$ with $N$ cycles. By inclusion-exclusion we have: $$d_{h,N} = \sum_{i=0}^N (-1)^i\binom hi c(h-i,N-i),$$ where $c(\cdot,\cdot)$ are (unsigned) Stirling numbers of 1st kind.

P.S. Numbers $d_{h,N}$ are called associated Stirling numbers of first kind and listed in the sequence A008306 in the OEIS.

The answer is $$\binom nh d_{h,N},$$ where $d_{h,N}$ is the number of derangements of size $h$ with $N$ cycles. By inclusion-exclusion we have: $$d_{h,N} = \sum_{i=0}^N (-1)^i\binom hi c(h-i,N-i),$$ where $c(\cdot,\cdot)$ are (unsigned) Stirling numbers of 1st kind.

The answer is $$\binom nh d_{h,N},$$ where $d_{h,N}$ is the number of derangements of size $h$ with $N$ cycles. By inclusion-exclusion we have: $$d_{h,N} = \sum_{i=0}^N (-1)^i\binom hi c(h-i,N-i),$$ where $c(\cdot,\cdot)$ are (unsigned) Stirling numbers of 1st kind.

P.S. Numbers $d_{h,N}$ are called associated Stirling numbers of first kind and listed in the sequence A008306 in the OEIS.

Source Link
Max Alekseyev
  • 34.3k
  • 5
  • 74
  • 152

The answer is $$\binom nh d_{h,N},$$ where $d_{h,N}$ is the number of derangements of size $h$ with $N$ cycles. By inclusion-exclusion we have: $$d_{h,N} = \sum_{i=0}^N (-1)^i\binom hi c(h-i,N-i),$$ where $c(\cdot,\cdot)$ are (unsigned) Stirling numbers of 1st kind.