Problem 210

To solve this problem, we can use a multi-step counting process.

First, we need to choose two balls to place in the boxes with their corresponding numbers. There are $\binom{6}{2} = 15$ ways to choose two balls out of six.

Next, we have four remaining balls and four boxes that need to be filled such that no ball is placed in a box with the same number. This is an example of a derangement problem for four items, which can be calculated using the derangement formula or by logical reasoning.

Let's fill the boxes in a sequence. The first ball has 3 choices (it can't go in its own box, and two boxes are already filled with the correct balls). The second ball also has 3 choices, taking into account the previous placement. Following this logic, the third ball would have 2 choices, and the last ball would have only 1 choice left.

However, the previous step didn't account for a potential issue where the third and fourth balls could still end up in their corresponding numbered boxes, which would violate the derangement requirement. To correct this, we need to subtract the cases where the last two balls are placed in their own boxes, which is 1 case (since if the third ball goes into its own box, the fourth ball has no choice but to do the same).

Effectively, for the last four balls, we have $(3 \times 3) - 1 = 9 - 1 = 8$ valid derangements.

Multiplying the two steps together, we find:
$$\text{Total number of ways} = \binom{6}{2} \times 8 = 15 \times 8 = 120$$

Therefore, there are $\boxed{120}$ ways to place the six balls into the boxes such that exactly two balls are placed into boxes with the same number.