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 option is 81.

Let us break down the solution step-by-step:

Step 1: Understand the grid layout
The problem states that 9 cups are arranged in an equal number of rows and columns. Since the total number of cups is 9, they must be arranged in a 3 × 3 grid, consisting of 3 rows and 3 columns. We have:
- Total cups = 9
- Cups containing coffee (C) = 6
- Cups containing tea (T) = 3

Step 2: Find the total number of unrestricted arrangements
The total number of ways to arrange 6 identical coffee cups and 3 identical tea cups in the 9 grid positions is given by the combination formula:

$\text{Total arrangements}=\left(\genfrac{}{}{0ex}{}{9}{3}\right)=\frac{9\times 8\times 7}{3\times 2\times 1}=84$

Step 3: Identify the invalid arrangements
We want to find the number of ways where each row contains at least one cup of coffee. The complement (invalid case) of this condition is that at least one row contains no coffee cups at all. Since each row has exactly 3 positions, a row with no coffee must be filled entirely with 3 tea cups.

Because there are only 3 tea cups in total, at most one row can consist entirely of tea. Thus, the invalid arrangements are those where exactly one of the rows contains all 3 tea cups. We can choose which row contains all tea cups in:

$\left(\genfrac{}{}{0ex}{}{3}{1}\right)=3\text{ways}$

Once a row is chosen to be filled with 3 tea cups, the remaining 6 positions in the other two rows must be filled with the 6 coffee cups, which can only be done in 1 way.

Therefore, there are exactly 3 invalid arrangements (where either Row 1, Row 2, or Row 3 contains only tea).

Step 4: Calculate the valid arrangements
Subtract the invalid arrangements from the total arrangements to get the number of ways each row has at least one cup of coffee:

$\text{Valid arrangements}=84-3=81$