- 31 Mar '07 16:20Last 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 - 31 Mar '07 16:25i 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? - 31 Mar '07 21:27 / 1 editBishops 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.