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9 years, 1 month ago
9 years, 1 month ago
Finding a:
Case 1: All the boxes have 1 ball each => 7! ways
Case 2: One box has 0 balls and one box has 2 balls, all the other boxes have 1 ball each => 7C1 (for selecting the box with 0) * 6C1 (selecting the box with 1)* 7C2 (selecting 2 balls to be placed in the box) * 5! (distribution of the remaining balls in the 5 boxes) ways
Case 3: Case of (2 2 1 1 1 0 0) => 7C2 (for selecting 2 boxes that have 0 balls) * 5C2 (for selecting 2 boxes that will have 2 balls each)* 7C2 (balls to be placed in the first box)* 5C2 (balls to be placed in the second box) * 3! (distribution of the remaining 3 balls) ways
Case 4: Case of (2 2 2 1 0 0 0) => 7C3 (selecting 3 boxes that will have 0 balls each) * 4C3 (selecting the box with 1 ball) * 7C2 (selecting 2 balls) * 5C2 (selecting the next set of 2 balls) * 3C2 (selecting the third set of 2 balls) ways
Sum of these = 463680
Finding b: There in only 1 way in which 7 distinct balls can be put in 7 similar boxes such that each box has only 1 ball
So, a/b = 463680
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