Simple way to calculate number of chess material combinations - algorithm

Question

In chess, one player can have different material combinations, for example:
"1 queen, 2 rooks, 2 knights, 2 bishops, 8 pawns + the king" is one combination
if the player loses one bishop:
"1 queen, 2 rooks, 2 knights, 1 bishop, 8 pawns + the king" is another combination
..afterwards, if a pawn is promoted to a knight, then:
"1 queen, 2 rooks, 3 knights, 1 bishop, 7 pawns + the king" is another combination
OK, the following combination is not valid:
"5 queens, 5 rooks, 5 knights, 5 bishops, 2 pawns + the king"
since you lack of pawns to promote. (5 queens = 4 pawns needed) (5 rooks = 3 pawns needed) , etc. so 4 + 3 + 3 + 3 = 13 pawns needed. Since 2 pawns on the board, then at most 6 pawns could be promoted. Not valid.
How many valid material combinations are there?
I computed 8694 combinations using the following C code. The question is:
Do you find simpler/efficient algorithm to calculate it? (less cycles, less calculations, clearer code, etc.) ... or even a math formulae??
total = 0;
for (queens=0;queens<=9;queens++)
for (rooks=0;rooks<=10;rooks++)
for (bishops=0;bishops<=10;bishops++)
for (knights=0;knights<=10;knights++)
for (pawns=0;pawns<=8;pawns++)
{
pawnsRequested = 0;
if (queens>1) pawnsRequested += queens - 1;
if (rooks>2) pawnsRequested += rooks - 2;
if (bishops>2) pawnsRequested += bishops - 2;
if (knights>2) pawnsRequested += knights - 2;
if (8-pawns < pawnsRequested) continue;
total++;
}
printf("%i\n",total);

Answer 1

If the piece types were independent, then we could just multiply: 10 possibilities for the queens times 11 possibilities for the rooks times etc. We need to track pawn usage, however. There's a mathematical trick called generating functions where we can encode the possibilities for, e.g., rooks as
3 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8,
where the power of x denotes the number of pawns used, and the coefficient denotes the number of possibilities. Here, there are three possibilities that require no promoted pawns (0, 1, 2), one that requires one promoted pawn (3), one that requires two promoted pawns (4), etc. Now we can multiply each of the factors together (respectively, queens, rooks, bishops, knights, pawns).
(2 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8)
* (3 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8)
* (3 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8)
* (3 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8)
* (1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8)
Here it is from Wolfram Alpha.
The coefficients of 1 through x^8, which are the number of possibilities for 0 to 8 pawns required, are 54, 135, 261, 443, 693, 1024, 1450, 1986, 2648, summing to 8694.