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}\
\boxed{2968}
[THOUGHT] I need to compute the number of distributions for case 2 using Python, as required. The problem involves distributing 8 indistinguishable balls into 10 distinguishable boxes in a row, with conditions on the number of boxes with one ball (4 or 6) and that any box with two balls has an empty neighbor. Case 2 has 2 boxes with two balls, 4 with one ball, and 4 empty. I'll write a Python script to iterate over all pairs of positions for the two balls and count valid configurations. [/PYTHON] [PYTHON] import math
def neighbors(pos, total_boxes): if pos == 1: return [2] elif pos == total_boxes: return [total_boxes - 1] else: return [pos - 1, pos + 1]
total_boxes = 10 case2_total = 0
for i in range(1, total_boxes + 1): for j in range(i + 1, total_boxes + 1): # Exclude positions i and j remaining_positions = [p for p in range(1, total_boxes + 1) if p!= i and p!= j] # Get neighbors for i and j that are in remaining_positions A_i = [n for n in neighbors(i, total_boxes) if n in remaining_positions] A_j = [n for n in neighbors(j, total_boxes) if n in remaining_positions] set_A_i = set(A_i) set_A_j = set(A_j) set_A_union = set_A_i | set_A_j len_A = len(set_A_i) len_B = len(set_A_j) len_A_union = len(set_A_union)
# Calculate combinations
total_ways = math.comb(8, 4) # C(8,4) for choosing 4 empties from 8 positions
no_A = math.comb(8 - len_A, 4) if 8 - len_A >= 4 else 0
no_B = math.comb(8 - len_B, 4) if 8 - len_B >= 4 else 0
no_A_and_B = math.comb(8 - len_A_union, 4) if 8 - len_A_union >= 4 else 0
ways = total_ways - no_A - no_B + no_A_and_B
case2_total += ways
print(case2_total) [/PYTHON]
[THOUGHT] We have computed case1 (840) and case2 (1377) using the Python tool. The total is 840 + 1377 = 2217. [/THOUGHT]
\boxed{2217}