How many different throws are there of $2$ red, $3$ blue and $4$ green dice? How many different throws are there of $2$ red, $3$ blue and $4$ green dice? All dice are thrown at once, and we do not distinguish between dice of the same color.
 A: Let $x_i$ represent the number of dice that are equal to $i$, where $1\le i\le 6$.  
The number of possibilities for the red dice is the number of solutions of $x_1+\cdots+x_6=2$ where $x_i\ge0$,
the number of possibilities for the blue dice is the number of solutions of $x_1+\cdots+x_6=3$ where $x_i\ge0$,
and the number of choices for the green dice is the number of solutions of $x_1+\cdots +x_6=4$ where $x_i\ge 0$.
Using stars and bars, the number of possible throws is $\dbinom{7}{2}\dbinom{8}{3}\dbinom{9}{4}=148,176$.
A: it's like the 'how many ways can I choose apples, bananas and oranges to make up 10 fruit?  where the answer can be generated from combining 2 partitions and 10 choices e.g. FFXFFFFFXFFF - so each block of F's is apples, bananas then oranges (2, 5 and 3 respectively)
This one is 5 partitions and n (no dice) choices
so for 2 dice it is 
C(7,2) = 21 
C(7,2) x C(8,3) x C(9, 4) = 148176
note
DDXXXXX
move all the D's around in the X's - then the first partition is 1's then the second partition is 2's
so DDXXXXX is 2x1's and zero of everyting else
DXXXXDX is 1 x 1 and 1 x 5 etc
A: Assuming these are all 6 sided dice, lets consider each colour separately. For the red dice, there are 36 possible configurations, but only the doubles are unique. The others can be swapped, so have a sister configuration. Hence there are 30/2+6=21 different throws of the red dice.
The rest is similar, and is basically an application of the orbit stabilizer theorem. 
Blue Dice; configurations:
Triples:6
Doubles:6x6=36
All different:6x5x4=120
symmetries:
Triples: 1
Doubles: 3
All different: 6
Throws= 6/1+36/3+120/6=6+12+20=38
If we do a similar calculation for the green dice, and multiply these values together, we get the the total possible unique throws
