The number of Latin rectangles [article]

Peter G. Doyle
2007 arXiv   pre-print
We show how to generate an expression for the number of k-line Latin rectangles for any k. The computational complexity of the resulting expression, as measured by the number of additions and multiplications required to evaluate it, is on the order of n^(2^(k-1)). These expressions generalize Ryser's formula for derangements.
arXiv:math/0703896v1 fatcat:w26zzme7dfaeldtl2yedhv63ka