Last night I was thinking about the game Chess960
It is stated that there are 960 possible piece settings and I was trying to think of a way to calculate that number.
Pieces have to be set from a few rules:
-king is placed between the two rooks
-pawns are on their normal rank
-bishops are placed on opposite-coloured squares
-black pieces are placed equal but, of course, mirrored
Can someone come up with a formula so you get to the number 960
i already came up with this:
King has to stay between rooks, so the other pieces have to be between the king and a rook or behind a rook. This leads to a mathematical combination:
8 nCr 5= 8!/(3!*5!)=56
Now pieces are put in these 'slots' and taken in account that bishops and knights are double, we multiply this by:
5!/(2!*2!*1) or 5 nCr 2 * 3 nCr 2 *1 nCr 1=30
Still this leads to a number far greater than 960, i.e.:1680
This because I've forgotten the bishops who have to be on different-coloured squares. But how do I do that?
Bishops have to be on opposite colored squares; each has four choices
4x4 = 16.
Six squares left for the queen:
16x6 = 96
Five squares left for knight #1:
96x5 = 480
Four squares left for knight #2:
480x4 = 1920
But in fact the knights are indistinguishable; if we switched them we wouldn't notice anything different. Divide by two to account for this symmetry:
1960/2 = 960.
Now there are three squares left for the king and rooks, and they must go in order from left to right: R-K-R.
So there are 960 possible starting positions.
This doesn't take into account left-right symmetry, so in fact half of these are simply mirror images of the other half and have no real differences between them.