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?

Nine cups in equal rows and columns means a 3 × 3 grid. We must place 3 tea cups so that no row is all tea; equivalently, each row must have at least one coffee.

• Option (a) 18: If tea distribution across rows is (2,1,0), choose the row with 2 teas in 3 ways, choose 2 positions in that row in 3 ways, and 1 position in another row in 2 × 3 = 6 ways. Total = 3 × 3 × 6 = 54, so 18 is incorrect.
• Option (b) 27: This does not match the valid count from row-wise cases. Incorrect.
• Option (c) 54: Valid tea placements are of two types: (1,1,1) gives 3^3 = 27 ways, and (2,1,0) gives 54 ways. Total valid arrangements = 27 + 54 = 81, so 54 alone is incorrect.
• Option (d) 81: Total ways to place 3 teas = (\binom{9}{3} = 84). Invalid cases are rows with all 3 teas in one row = 3 ways. Hence valid arrangements = 84 - 3 = 81. Correct.