There are 9 cups placed on a table arranged in equal number of rows and columns out of which 6 cups contain coffee and 3 cups contain tea. In how many ways can they be arranged so that each row should contain at least one cup of coffee?
18
27
54
81
The cups will be arranged in a 3 × 3 matrix. We need to remove cases where any one row contains only Tea ie all 3 Teas in a single row. This can happen in 3 ways - first, second or third row. Therefore 3 cases need to subtracted from total combinations possible. Total arrangements = 9! /(6! x 3!) (by formula) Required answer = total arrangements possible - 3 = 84 - 3 = 81
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