A square is divided into 9 identical smaller squares. Six identical balls are to be placed in these smaller squares such that each of the three rows gets at least one ball (one ball in one square only). In how many different ways can this be done?

To find the total number of ways to place 6 identical balls in 9 squares such that each row has 3 squares and receives at least one ball, we follow these steps:

1. Total combinations: First, calculate the total ways to choose 6 squares out of 9 without any restrictions. This is 9C6, which is equal to 9C3. Calculation: (9 x 8 x 7) / (3 x 2 x 1) = 84 ways.

2. Identify restricted cases: We must subtract the cases where one or more rows remain empty. Since there are 6 balls to be placed in 9 squares, it is impossible for two rows to be empty because one row only has 3 squares, and we have 6 balls. Therefore, we only need to subtract cases where exactly one row is empty.

3. Calculate empty row cases: If one row is empty, all 6 balls must be placed in the remaining two rows. Since the remaining two rows have exactly 6 squares combined, there is only 1 way to fill them (1 ball in each square).

4. Apply the subtraction: There are 3 possible rows that could be the empty one (Row 1, Row 2, or Row 3). So, there are 3 such cases where a row remains empty.

5. Final result: Total ways minus restricted ways = 84 minus 3 = 81.

Thus, there are 81 different ways to place the balls such that each row gets at least one ball. Correct option is D.