To solve this problem, let's consider the placement of the balls in the smaller squares. Since each row should have at least one ball, we can divide the problem into three cases:
Case 1: Each row has exactly one ball.
In this case, we have one ball to place in each row. The number of ways to do this is equal to the number of ways to arrange the balls in the rows, which is 3!.
Case 2: One row has two balls, and the other two rows have one ball each.
In this case, we need to choose one row to have two balls and then place the remaining ball in one of the two remaining rows. The number of ways to do this is equal to the number of ways to choose the row (3) multiplied by the number of ways to choose the remaining row (2).
Case 3: One row has three balls, and the other two rows have one ball each.
In this case, we need to choose one row to have three balls and then place the remaining balls in the two remaining rows. The number of ways to do this is equal to the number of ways to choose the row (3).
Now, let's calculate the total number of ways by summing up the possibilities from each case:
Total number of ways = (number of ways in Case 1) + (number of ways in Case 2) + (number of ways in Case 3)
= 3! + (3 * 2) + 3
= 6 + 6 + 3
= 15
Therefore, there are 15 different ways to place the balls in the smaller squares.
The correct answer is A) 81.