In how many different ways can 5 different toys be packed into 3 identical boxes such that no box is left empty and any box may hold any number of toys?

Introduction / Context:
This question involves distributing distinct objects (toys) into identical boxes with a non emptiness condition. When boxes are identical, we cannot distinguish between arrangements that differ only by swapping the boxes. Counting such configurations is related to set partitions and Stirling numbers of the second kind. For 5 distinct toys and 3 identical non empty boxes, the answer equals the Stirling number S(5, 3).

Given Data / Assumptions:
Number of toys = 5. Each toy is different. Number of boxes = 3. Boxes are identical (not labeled), so their order does not matter. No box may remain empty. Any box may hold one or more toys, as long as the total toys remain 5.

Concept / Approach:
We are essentially partitioning a set of 5 distinct elements into exactly 3 non empty unlabeled groups. The number of such partitions is the Stirling number of the second kind S(5, 3). This Stirling number can be computed using a recurrence relation, a combinatorial reasoning approach, or a direct formula. For small values, it is also well known and tabulated: S(5, 3) equals 25. Because the boxes are identical, we do not distinguish permutations of the groups themselves.

Step-by-Step Solution:
Step 1: Recognize that distributing toys into 3 identical boxes with no empty box corresponds to partitions of the set of 5 toys into 3 non empty unlabeled subsets. Step 2: Use the Stirling number of the second kind S(n, k) to count partitions of an n element set into k non empty unlabeled subsets. Step 3: Here, n = 5 and k = 3, so we need S(5, 3). Step 4: Using the recurrence relation S(n, k) = S(n - 1, k - 1) + k * S(n - 1, k), along with known base values, we can compute S(5, 3). Step 5: Known values include S(3, 1) = 1, S(3, 2) = 3, S(3, 3) = 1, S(4, 1) = 1, S(4, 2) = 7, S(4, 3) = 6 and S(4, 4) = 1. Step 6: Compute S(5, 3) = S(4, 2) + 3 * S(4, 3) = 7 + 3 * 6 = 7 + 18 = 25. Step 7: Hence the number of distinct packing patterns is 25.

Verification / Alternative check:
You can also reason combinatorially about possible distributions of sizes. Since boxes are non empty and identical, the group sizes of toys can only be 3-1-1 or 2-2-1. One may count how many ways to choose which toys go into each group size arrangement, then divide out the permutations of groups of the same size to respect identical boxes. Summing over 3-1-1 and 2-2-1 distributions also leads to 25. This more detailed method is more tedious but matches the Stirling number result, confirming the answer.

Why Other Options Are Wrong:
The value 36 might arise from treating boxes as distinct or from using a combinations and permutations mix incorrectly. The value 24 often comes from simple factorial thinking like 4!, which does not model the problem structure. The value 72 is far too large for unlabeled boxes and mixes labeled box counting with non emptiness constraints. Only 25 aligns with the correct set partition interpretation using S(5, 3).

Common Pitfalls:
The most common error is forgetting that the boxes are identical, not labeled. Many students first count distributions into labeled boxes and then forget to divide out the permutations of boxes that represent the same grouping. Another pitfall is allowing empty boxes, which would correspond to partitions into at most 3 subsets instead of exactly 3. Some also confuse this with distributing identical objects, which is a completely different problem. Always check whether objects or boxes are distinct and whether emptiness is allowed before choosing the counting method.

Final Answer:
The number of different ways to pack the 5 toys into 3 identical non empty boxes is 25.