How many ways we can split $7$ green balls , $9$ red balls, and $10$ yellow balls to $2$ equal groups I want to calculate how many ways we can split $7$ green balls , $9$ red balls, and $10$ yellow balls to $2$ equal groups. I want to check 2 options:


*

*The order is importent

*The order is not importent


How I can think about that? if the order is not importent I can refer is to "throw" 26 balls on 2 places, $26!$ is all the ways, but we have the same balls so its need to be $\frac{26}{9!\cdot 7! \cdot 10!}$ and put this on 2 places so I can use Binominal $p(2,\frac{26}{9!\cdot 7! \cdot 10!}$) 
what about the first one? I need to use generating functions?
thanks!
 A: Case 1
Because the order is important we can give the balls a numbers form 1-26. At the start there will be 26 balls to chose, then 25, then 24 ... ath the end there will be 13 to chose. So the number of permutation is:
$$\frac{n!}{(n-k)!} = \frac{26!}{13!}\text{ ways to choose}$$
Note that if we take the balls with numbers from 1-13 and 14-26 is completely the same, because in the first case the balls from 1-13 will be in group A and balls from 13-26 in group B, and in the second case vice versa.
Case 2 
In this case the order is not important so the problem is to pick 13 balls from 3 sets of limited supply of balls.
We can choose $(1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7)$ green balls.
We can choose $(1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8 + x^9)$ red balls.
And at last we can chose $(1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8 + x^9 + x^{10})$ yellow balls.
Multiply these three polynomials and look at the coefficient in front of $x^{13}$, beacuse this is the number of balls we want to choose.
We find that the coefficient in front of $x^{13}$ is 68, so there are 68 ways to chose 13 balls from 3 sets of limited supply of balls.
But because every set of 13 balls has a complementary set we must divide the number by 2.
Here's what I mean. For example we chose
(3 green, 5 red, 5 yellow), which is the same as choosing (4 green, 4 red, 5 yellow), because if we choose the first set to be group A, then the second will be group B. And if we choose second set to be group B, then the first will be group A, which is the same.
So there are 34 distinct cases to select 13 balls when the order isn't important.
