Let's say we have a full Chess set - 2 Kings (1W, 1B), 4 Knights (2W, 2B), 16 Pawns and so on.
How many ways can we arrange just half a set by picking pieces of either color. One example arrangement can be:
| a | b | c | d | e | f | g | h |
|---|---|---|---|---|---|---|---|
| BP | WP | WP | BP | BP | WP | WP | WP |
| WR | BN | BB | BQ | WK | WB | WN | WR |
Legend:
W : White | B : Black
R : Rook | B : Bishop | N : Knight | K : King | Q : Queen
WR : White Rook | BB : Black Bishop
Thanks for the help!
On
So we're working on 32 cases of the chessboard. Let's consider the BK. We have 32 cases where we could place him. Now let's consider the WK. There is 31 cases. We keep going, and finally we find 32!. But that's not right, because there is 2 identics Knights, bishops and rooks for each colour, and 8 pawns, so we have to divide the results by 2^6 x 2^6 = 2^12, so i'd say that the answer is 32!/2^12.
On
The following solution uses an exponential generating function.
For those who don't know chess, each of the two sides (White and Black) has $1$ King, $1$ Queen, $2$ Bishops, $2$ Knights, $2$ Rooks, and $8$ Pawns, for a total of $16$ pieces. I'm assuming that pieces of the same type and color are indistinguishable. We want to find the number of distinct permutations taken from a collection of $16$ pieces. More generally, we might ask what is the number of permutations of $n$ pieces where $0 \le n \le 32$. The exponential generating function of the number of permutations is $$f(x) = (1+x)^4 \left(1 + x + \frac{1}{2!} x^2 \right)^6 \left(1 + x + \frac{1}{2!} x^2 + \frac{1}{3!} x^3 + \dots + \frac{1}{8!} x^8 \right)^2$$ The number of permutations of $n$ pieces is $n! [x^n]f(x)$, i.e. $n!$ times the coefficient of $x^n$ when $f(x)$ is expanded. Expanding $f(x)$ would clearly be tedious if done by pencil and paper, but it's easy if you use a computer algebra system. In the case $n=16$, we find $$16! [x^{16}]f(x) = 2\;247\;426\;233\;099\;190$$
This solution takes the assumption that two pieces of the same kind and color are indistinguishable from one another
As correctly pointed out by someone in the comments below, we are simply decided the number of white pieces that there will be (or black, but it doesn't change the answer) and we can then pick anywhere from 0 to the maximum number of pieces of that kind being white which gives us a total of the number of that type of piece to be chosen +1 different ways to pick each piece.
King: 2
Queen: 2
Bishops: 3
Knights: 3
Rooks: 3
Pawns: 9
This gives us a total of $2^23^39$ = 972 ways