JEE Advanced 2022 Paper 2 · Q15 · Combinations

Question

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?

SolveKar AI · Accepted Answer

Per box, the number of (red, blue) selections with at least one of each must be tracked. Across 4 boxes, the total red+blue chosen $=10$. Let $r_i\in\{1,2,3\}$ red from box $i$ (since at least 1 red, at most 3) and $b_i\in\{1,2\}$ blue. Constraint $\sum(r_i+b_i)=10$, with each $r_i+b_i\in\{2,3,4,5\}$. Using generating function approach: number of ways = coefficient of $z^{10}$ in $\prod_{i=1}^4 [(\binom{3}{1}z + \binom{3}{2}z^2+\binom{3}{3}z^3)(\binom{2}{1}z+\binom{2}{2}z^2)] = (3z+3z^2+z^3)^4(2z+z^2)^4$. After full expansion and extraction, the coefficient of $z^{10}$ yields $21816$. Hence option (A).

JEE Advanced 2022 Paper 2 · Q15 · Combinations

Solution