Consider $4$ boxes, where each box contains $3$ red balls and $2$ blue balls. Assume that all $20$ balls are distinct. In how many different ways can $10$ balls be chosen from these $4$ boxes so that from each box at least one red ball and one blue ball are chosen?
(A) 21816
(B) 85536
(C) 12096
(D) 156816
My Attempt
I tried to take different cases
(2,2)(1,1)(1,1)(1,1) $\times 4$
(1,2)(2,1)(1,1)(1,1) $\times 6\times 2$
(3,1)(3,1)(1,1)(1,1) $\times 6$
(1,2)(1,2)(1,1)(1,1) $\times 6$
(R,B) denotes number of red balls and blue balls chosen from the bag.
But it appears I am missing certain cases
Best Answer
As we need to choose at least one red and blue ball out of every box, $8$ of $10$ places are already chosen.
I come across the following five possibilities (noted similarly with the R,B-pairs):
$$a: (3,1),(1,1),(1,1),(1,1) \\ b: (2,1),(2,1),(1,1),(1,1) \\ c: (2,1),(1,2),(1,1),(1,1) \\ d: (2,2),(1,1),(1,1),(1,1) \\ e: (1,2),(1,2),(1,1),(1,1)$$
Now you need to consider that all balls are distinct. This means $(1,1)$ actually represents $3\cdot 2$ and $(2,2)$ represents $3\cdot 1$ possibilities! You can also swap the boxes with each other, so for example in case $a$ by swapping the boxes we need to multiply everything by $4$ resulting in the following number of possibilities:
$$ a: (2 \cdot 6 \cdot 6 \cdot 6) \cdot 4 \\ b: (6 \cdot 6 \cdot 6 \cdot 6) \cdot 6 \\ c: (6 \cdot 3 \cdot 6 \cdot 6) \cdot 12 \\ d: (3 \cdot 6 \cdot 6 \cdot 6) \cdot 4 \\ e: (3 \cdot 3 \cdot 6 \cdot 6) \cdot 6 \\ $$ resulting in a total of $1728+7776+7776+2592+1944=21816$ cases.
This means $(A)$ is correct.