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
Let the four boxes be Box-1, Box-2, Box-3 and Box-4. Each of the box has 3 red balls and 2 blue balls. 10 balls can be chosen in: Case I - When exactly one box gives four balls Case II - When exactly two boxes gives three balls Required number of ways = Number of ways in Case I + Number of ways in Case II. = C 1 4 C 1 2 + C 1 3 C 1 3 C 1 2 3 + C 2 4 C 2 3 C 1 2 + C 1 3 C 2 2 2 C 1 3 C 1 2 2 = 4 5 6 3 + 6 3 × 2 + 3 2 6 2 = 4320 + 17496 = 21816
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