Consider $4$ boxes, where each box contains $3$ red balls and $2$ blue balls. In how many different ways can $10$ balls be selected from these four boxes so that from each box at least one red ball and at least one blue ball are chosen?
(A) $21816$
(B) $85536$
(C) $12096$
(D) $156816$
First, I go to one of the four boxes and select a red ball and a blue ball which I would be able to do in $3\times2$ ways or $6$ ways. Then I continue the process for all four boxes resulting in $6^4$ ways of selecting at least one red ball and one blue ball from all the four boxes. Now $8$ of the ten balls required haven been selected and each box now contains $2$ red balls and $1$ blue ball, making a total of $12$ unselected balls. Since all the $12$ balls are distinct, the number of ways of selecting the remaining $2$ required balls should be $12\choose2$. Therefore the final answer must be ${6^4}\times{12\choose2}$ which is $85536$. But the correct answer is apparently the $21816$. Where exactly did I go wrong?
Best Answer
Note your method gives repeated countings. We take the first bag for example, and we mark the three red balls as $r1, r2, r3$, two blue balls as $b1, b2$. If assume your last two balls are drawn from the first bag, you may have:
$$r1, b1, r2, r3$$
$$r2, b1, r1, r3$$
These are repeated countings, hence your answer is greater than the correct answer.
To avoid repeated countings, you may discuss it in two different cases:
Case.(1) Ten balls are distributed as $4, 2,2,2 $ ($4$ can be from the first, second, third , fourth bag)
$$4\cdot \left[\binom{5}{4}\cdot 6^3\right]=4320$$
Case.(2) Ten balls are distributed as $3,3,2,2$
$$\binom{4}{2}\cdot \left[\binom{5}{3}-1\right]^2\cdot 6^2=17496$$
Totally,
$$4320+17496=21816$$