Permutations without repeats

61 Views Asked by At

Consider this matrix:

A B C D E
D C A E B
B A E C D
E D B A C
C E D B A

The letters are arranged so that no row and no coulumn contains the same letter twice (Sudoku style).

Let's call the number of diffent letters $n$ (5 in the above example). When writing the first row, I have $n!$ permutations to choose from.

How many permutations can I choose between when writing the second row? And the third row? Etc.

1

There are 1 best solutions below

0
On BEST ANSWER

When entering the second row, it has to be a derangement of the first row. When $n=5$ there are $44$ derangements. But from the third row onwards the number of possibilities will depend on exactly what the previous rows contain.