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 ?

$\text{Each box: } 3\ \text{red} + 2\ \text{blue} = 5 \text{ balls}$ $4\ \text{boxes} \Rightarrow 12\ \text{red} + 8\ \text{blue} = 20\ \text{distinct balls}$ $\text{Choose } 10 \text{ balls, } \ge 1 \text{ red and } \ge 1 \text{ blue from each box}$ $\text{To find: number of ways to choose the } 10 \text{ balls}$ $\text{Hint: each box gives at least } 2 \text{ balls } (1R + 1B)$ $\text{Only } 2 \text{ blue exist per box} \Rightarrow b \in \{0,1,2\} \text{ per box}$ $\text{Each box: } \ge 1R,\ \ge 1B \Rightarrow \ge 2 \text{ balls}$ $4 \text{ boxes} \Rightarrow 8 \text{ mandatory balls}$ $10 - 8 = 2 \text{ extra balls to distribute}$ $\text{Plan: (i) per-box ways for } e = 0,1,2,3 \text{ extras}$ $\text{(ii) distribute the } 2 \text{ extras over } 4 \text{ boxes}$ $\text{(iii) add the cases}$ $r \in \{1,2,3\}, \quad b \in \{1,2\}$ $\text{Ways from one box} = \binom{3}{r}\binom{2}{b}$ $e = (r + b) - 2 \text{ extras}$ $e=0:\ \binom{3}{1}\binom{2}{1} = 3 \cdot 2 = 6$ $e=1:\ \binom{3}{2}\binom{2}{1} + \binom{3}{1}\binom{2}{2} = 6 + 3 = 9$ $e=2:\ \binom{3}{2}\binom{2}{2} + \binom{3}{3}\binom{2}{1} = 3 + 2 = 5$ $e=3:\ \binom{3}{3}\binom{2}{2} = 1$ $\text{Per-box: } 6 + 9y + 5y^2 + y^3$ $N = [y^2]\,(6 + 9y + 5y^2 + y^3)^4$ $\text{2 extras } \Rightarrow \text{ two cases}$ $\text{Case A: one box } e=2, \text{ the other three } e=0$ $4 \cdot 5 \cdot 6^3 = 4 \cdot 5 \cdot 216 = 4320$ $\text{Case B: two boxes } e=1, \text{ the other two } e=0$ $\binom{4}{2} \cdot 9^2 \cdot 6^2 = 6 \cdot 81 \cdot 36 = 17496$ $N = 4320 + 17496 = 21816$ $\text{Answer: } (1)\ 21816$