Ten empty boxes are arranged in a row, and there are eight balls. The balls are to be placed into the boxes so that no boxes are left unused, the total number of balls is , and each box contains at most two balls. (The balls are indistinguishable.) Find the number of such distributions that satisfy the following conditions. [4 points]\begin{center}\fbox{\begin{minipage}{0.9\textwidth}(A) The number of boxes containing exactly one ball is either or .(B) Any box containing two balls has empty neighboring boxes.\end{minipage}}\end{center}\
[THOUGHT] The problem involves distributing 8 indistinguishable balls into 10 distinguishable boxes under certain conditions. We need to find the number of such distributions that satisfy two conditions: (A) The number of single-ball boxes is either 4 or 6. (B) Any box containing two balls must have empty neighboring boxes.
Let's start by considering Condition (B). If a box contains two balls, it must be flanked by empty boxes. This means we can't place any more than one ball in the adjacent boxes. Therefore, each pair of consecutive boxes with two balls between them effectively counts as a single 'unit'.
Now let's look at Condition (A) and consider the possible numbers of such units (single-ball boxes). We have 10 boxes in total, so we can create a maximum of 9 units (since each unit consists of 3 boxes). However, our task is to find distributions where the number of single-ball boxes is either 4 or 6.
If there are 4 single-ball boxes, it means there are 6 units (each unit being 2 boxes + 1 empty box). Similarly, if there are 6 single-ball boxes, we can create 4 units.
So, our task boils down to finding the number of ways to distribute 8 identical balls into 6 or 4 'units', where each unit consists of 3 consecutive boxes (2 filled and 1 empty).
For 6 units:
- Choose 4 units out of 9: C(9,4) = 126
- In each unit, choose 1 ball to place in the first box (leaving 2 for the other two boxes): 4P1 = 4
- The remaining 4 balls are distributed among the 6 empty spaces (including the ends): This is equivalent to placing 4 identical items into 5 distinct bins, which can be solved using stars and bars method as C(6+4,4) = C(10,4) = 210
For 4 units:
- Choose 4 units out of 9: C(9,4) = 126
- In each unit, choose 2 balls to place in the first box (leaving 1 for the second and third boxes): 4C2 * 3P2 = 36
- The remaining 4 balls are distributed among the 5 empty spaces: C(5+4,4) = C(9,4) = 126
Total distributions: = (6 units case) + (4 units case) = (126 * 4 * 210) + (126 * 36 * 126) = 106280 + 508956 = 615236
So the final answer is:
\boxed{615236}