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
Each box must contribute at least one red and one blue ball. Each box has 3 red and 2 blue (5 distinct balls), and 10 balls are chosen in total from 20.
Distribute 10 balls across 4 boxes with each box giving balls (one red, one blue). Two distributions are possible:
Case 1: One box gives 4 balls, the other three each give 2.
Box of 4: choose at least 1 red and at least 1 blue out of 5 with 4 chosen. The only way is missing one of the 5 balls; valid selections are those missing a red (3 ways, leaving 2R+2B) or missing a blue (2 ways, leaving 3R+1B). Total 5 ways. Box-choice: .
Each box of 2: ways. Three such boxes contribute .
Case 1 total: .
Case 2: Two boxes give 3 balls each, the other two give 2 each.
Box of 3 (with at least 1R and 1B): either 2R+1B () or 1R+2B (), total 9 ways.
Choose the two 3-ball boxes: . Each 3-box: ; each 2-box: .
Case 2 total: .
Grand total: .
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