The number of ways to put 9 identical balls in 3 labelled boxes in such a way that each box contains at most 7 balls, is
将 9 个相同的球放入 3 个贴有标签的盒子中，使每个盒子最多包含 7 个球的方法数为

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BlackTom AI · Accepted Answer

We restate the problem: count the number of ways to put 9 identical balls into 3 labeled boxes with each box containing at most 7 balls. Without any cap, the number of nonnegative integer solutions to x1 + x2 + x3 = 9 is C(11,2) = 55.
Next, we enforce the cap by excluding the distributions where any box has 8 or more balls. For a fixed box i to have at least 8 balls, write xi = yi + 8 with yi >= 0. Then yi + xj + xk = 1, where j and k are the other two boxes. The number of nonnegative solutions to a + b + c = 1 is C(3,2) = 3. Since there are 3 choices for which box exceeds 7, we subtract 3 × 3 = 9 from 55, giving 46.
We should also check double or triple overlaps (two or all boxes >= 8). If two boxes were at least 8, the total would be at least 16, which is impossible since the total balls are 9. Hence there are no overlaps to add back in. Therefore, the final count comes from 55 minus 9, with no further adjustments.
Because the problem statement mentions a single numeric answer, and the calculation above yields the resulting count after applying the constraint, this is the derived total for the described scenario. Note that the provided answer options field is empty, so no multiple-choice alternatives are shown here. The logic hinges on first counting all distributions, then subtracting the infeasible ones due to the at-most-7 constraint, and confirming that higher-overflow overlaps cannot occur in this total.

MATH2069 (ND) MATH2069 （ND） Combinatorics Quiz 组合学测验

The number of ways to put 9 identical balls in 3 labelled boxes in such a way that each box contains at most 7 balls, is 将 9 个相同的球放入 3 个贴有标签的盒子中，使每个盒子最多包含 7 个球的方法数为

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