In how many ways can 11 identical toys be placed in 3 distinct boxes such that…

Concept: To distribute n identical items into k distinct groups so that every group receives at least one item, first reserve one item for each group. If x1, x2, ..., xk (each ≥ 1) denote the items in the groups with x1 + x2 + ... + xk = n, the number of positive-integer solutions equals C(n − 1, k − 1) by the Stars and Bars theorem.

Application:

1. Let x1, x2, x3 be the number of toys in the three boxes, so x1 + x2 + x3 = 11, with each xi ≥ 1 (no box may be empty).

2. Substitute yi = xi − 1, so each yi ≥ 0. The equation becomes y1 + y2 + y3 = 11 − 3 = 8.

3. The number of non-negative integer solutions to y1 + y2 + y3 = 8 is C(8 + 3 − 1, 3 − 1) = C(10, 2).

4. Compute C(10, 2) = (10 × 9) / (2 × 1) = 45.

Cross-check: Applying the Stars and Bars formula directly with n = 11 and k = 3 gives C(n − 1, k − 1) = C(10, 2) = 45 — the same result, confirming the answer.

So there are 45 ways to place the 11 identical toys into the 3 distinct boxes with none left empty.