Hi
I figured out a simple method a few years ago, first proof then in practice.
Proof
Only use 3 squares: A1, B1 and C1
Only use 3 pieces: K, Q, R
You have 3 squares and 3 pieces so the total permutations is 3 x 2 x 1 = 6.
Now try K + R + R
The actual permutations is still 6 like above but in chess the 2 rooks are considered the same.
So what you have to do is divide by the number of pieces as well like this.
K = 3 squares / One king = 3
R = 2 squares / Two rooks = 1
R = 1 square / One Rook = 1
Total permutations for K + R + R = 3 x 1 x 1 = 3
Note that this logic also works for pawns as each is considered the same = 8 squares / 8 pawns = 1.
Last thing is that the Bishops are NOT considered the same: Dark squares vs. Light squares.
Practice
You now just calculate the same way that you set up the pieces with a dice, same order below:
1. Light Bishop = 4 squares / One Bishop (Light) = 4
2. Dark Bishop = 4 squares / One Bishop (Dark) = 4
3. Queen = 6 squares / One queen = 6
4. Knight = 5 squares / Two Knights = 2.50
5. Knight = 4 squares / One Knight = 4
You now have only 3 squares left, there is ONLY one way to do it, King on square between the other two squares then place the rooks.
Answer: 4 x 4 x 6 x 2.50 x 4 x 1 = 960
I haven't been on this site for a while, wanted to share something (but wasn't sure where to post it, so sorry if this is the wrong place)
I figured out a way to calculate the positions in Chess960 (and by doing so confirm that it is 960) I'm pretty sure there are other ways to do it, but I just wanted to share the way that I discovered (I hope I've explained it well enough, if I haven't I can clarify it better, just comment below)
You start off with the "rook rule" (the rooks must be either side of the king) this means that the king is between the rooks, which means there is a least 1 square between the rooks and given that there is 8 squares in a row, there's 6 at most (the other two are occupied by the rooks)
Now if there are:
6 squares between the rooks, there is 1 way to position them (in the corners)
5 squares between the rooks, there is 2 ways to position them (you can move them along 1 square once)
4 squares between the rooks, there is 3 ways to position them (you can move them along 1 square twice)
3 squares between the rooks, there is 4 ways to position them (you can move them along 1 square three times)
2 squares between the rooks, there is 5 ways to position them (you can move them along 1 square four times)
1 square between the rooks, there is 6 ways to position them (you can move them along 1 square five times)
the number of squares between the rooks is the number of possible squares for the king so multiply the number of ways the rooks can be positioned by how many squares between them:
1 (way to position them) X 6 (squares between them)
2 (ways to position them) X 5 (squares between them)
3 (ways to position them) X 4 (squares between them)
4 (ways to position them) X 3 (squares between them)
5 (ways to position them) X 2 (squares between them)
6 (ways to position them) X 1 (square between them)
which is:
6X1 +
5X2 +
4X3
all doubled since they repeat
6+10+12 = 28 doubled is 56
now the bishops must go on opposite colours
now some of those 56 ways of positioning the king and rooks have all 3 on the same colour square so there is 1 square for one of the bishops and 4 for the other (4X1=4 possibilities)
for the others two of them (the rooks and the king) are on one colour the other (which could be the king or a rook) are on the other colour so there are 2 squares for one bishop and 3 for the other (2X3=6 possibilities)
of the 56 possible arrangements of the rooks and king
if there is an even number of squares between the rooks then they must be on opposite colour, so what the king is on is irrelevant (it will always be 2 of one and 1 of the other)
if there is 1 square between them, there's one square for the king which is the opposite colour (it will always be 2 of one and 1 of the other)
if there are 3 squares between them, there is one square between them of the same colour and 4 ways for the rooks to be positioned (as mentioned above) so its 4X1 =4 (4 of the 56 with the king and rooks on the same colour)
if there are 5 squares between them, there are 2 squares between them of the same colour and 2 ways for the rooks to be positioned (as mentioned above) so its 2X2 =4 (4 of the 56 with the king and rooks on the same colour)
a total of 8 of the 56 (4+4) have the king and rooks on the same colour so 4X8 (4 ways for the bishops to be positioned 8 ways for the rooks and king to be placed on the same colour)
56-8 = 48 (48 ways they are on at least two different colours) this means 48X6 (6 ways to position the bishops in this situation as shown above) 48X6 = 288
32+288 = 320 (ways to position the rooks, bishops and king)
there are 3 pieces left: the queen and the two knights
the queen can go:
1) the right the knights
2) the left of the knights
3) the middle of the knights
with 3 possibilities times 320 for the other pieces:
3X320 = 960