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?

Correct Answer: (D) 81

Placing 6 coffee cups and 3 tea cups in a $3$$\times$ 3 grid corresponds to choosing 3 positions out of 9 for the tea cups, giving $\left(\genfrac{}{}{0px}{}{9}{3}\right)=\frac{9\times 8\times 7}{6}=84$$\binom{9}{3}$ = $\frac{9 \times 8 \times 7}{6}$ = 84 total possible arrangements. The condition requires each of the 3 rows to contain at least one cup of coffee. The complement occurs when an entire row contains no coffee cups, meaning all 3 cups in that row are tea cups. Since there are exactly 3 tea cups, this can happen in only 3 ways (all 3 tea cups in Row 1, Row 2, or Row 3). Subtracting these 3 invalid configurations yields $84 - 3 = 81$ valid arrangements.