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?

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