Arrange 5 non-attacking knights on a 5x5 toroidal board The celebrated British mathematician and lover of public transport, Simon Norton, was one of those who passed in 2019. He was the subject of a wonderful biography and one of my favourite human beings: an amazing and innocent man. 

He was not a particular fan of chess, but here is a tiny gem due to him. 


How many ways are there to arrange 5 knights on a 5x5 toroidal chessboard so that none of them attack one another?


“Toroidal” means the board has no edges. You can leave on the left and just appear in the board on the right. You can leave at the top and just appear at the bottom. And vice versa.



EDIT: Jörg P. has totally nailed it, so I have to give him the green tick. But his solution doesn't give any insight as to why the number is what it is. So I guess that's a bonus question: 


Bonus question: why do you get this number (120)?
I noticed a simplification of @RosieF's solution:


  Label the squares with two-number labels as follows:


4,2 0,3 1,4 2,0 3,1
1,0 2,1 3,2 4,3 0,4
3,3 4,4 0,0 1,1 2,2
0,1 1,2 2,3 3,4 4,0
2,4 3,0 4,1 0,2 1,3

  
  Then two knights may occupy two squares x,y
  and X,Y iff those squares' first numbers x and X are different and
  their second numbers y and Y are different.


Since there are 5 knights and only 5 different numbers in each direction, that means that every x-value must be occupied by a knight, and each of the possible y-values must be assigned to one of those knights.  If we sort the knights' coordinates by their x-value, then list only the y-value, then we wind up with some permutation of the numbers from 0 to 4.  Every valid placement of the knights produces a permutation of the numbers 0-4, and every permutation of those numbers corresponds to a valid arrangement of the knights.

The total number of valid arrangements of the knights is thus 5!, or 120.