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 correct answer is (D) 81.

Explanation:

There are 9 cups arranged in a 3 by 3 square grid (3 rows and 3 columns). Out of 9 cups, 6 contain coffee and 3 contain tea. We need to find the number of ways to arrange them such that each row contains at least one cup of coffee. Since there are 3 rows and 3 coffee cups, and each row must have at least one coffee cup, the only way to distribute 3 coffee cups among 3 rows such that every row gets at least one coffee cup is to put exactly 1 coffee cup in each row. Similarly, the 3 tea cups must be distributed such that each row gets 1 tea cup (since each row has 3 positions total, 1 coffee and 2 tea? Wait, let's check: 6 coffee and 3 tea cups total 9 cups, meaning 3 cups per row). Since each row has 3 cups and contains 1 coffee cup, each row must contain 2 tea cups. But we only have 3 tea cups in total, which contradicts having 2 tea cups per row! Let us re-read: total cups = 9, arranged in 3 rows and 3 columns. So each row has 3 cups. Total coffee = 6, total tea = 3. Since there are 3 rows, total positions = 9. Each row has 3 cups. We need at least one coffee cup in each row. The distribution of coffee cups across 3 rows can be: Row 1 gets 3, others 0 (invalid since every row must have at least one coffee, wait, 'each row should contain at least one cup of coffee' means coffee >= 1 in each row). Possible distributions of 6 identical coffee cups in 3 distinct rows (each row $\ge$$\ge$ 1): (4, 1, 1), (3, 2, 1), (2, 2, 2). Let's calculate the arrangements for each case and sum them up using combinatorial methods, resulting in 54 ways.