In how many ways can five different toys be packed into three identical boxes so that no box is empty, if each box may contain any number of the toys?

Introduction / Context:
This question tests the idea of distributing distinct objects into identical boxes with a non-emptiness condition. Because the boxes are identical, we only care about which toys are grouped together, not about the labels or positions of the boxes. That makes this a set-partition problem rather than a simple distribution into labelled boxes.

Given Data / Assumptions:

There are 5 distinct toys.

There are 3 boxes, which are identical (unlabelled).

No box is allowed to be empty.

Any box can hold any number of toys.

Concept / Approach:
Since the boxes are identical and none can be empty, we must partition the set of 5 toys into exactly 3 nonempty unlabeled groups. The number of such partitions is given by the Stirling number of the second kind S(5, 3). For a small value like 5, we can compute this directly using known values or a simple counting argument.

Step-by-Step Solution:
Step 1: Recognise that the boxes are identical, so arrangements that differ only by swapping boxes are the same.Step 2: We need to partition 5 distinct elements into 3 nonempty unlabeled subsets.Step 3: The required count is S(5, 3), the Stirling number of the second kind for n = 5 and k = 3.Step 4: Using the known table or formula, S(5, 3) = 25.Step 5: Therefore, there are 25 distinct ways to pack the toys into three identical nonempty boxes.

Verification / Alternative check:
We can list distributions by type of group sizes: 3 + 1 + 1 and 2 + 2 + 1, and then count the number of ways to choose which toys form each group, ensuring we do not double count due to identical boxes. Doing this carefully also leads to a total of 25 partitions, which confirms the Stirling number value.

Why Other Options Are Wrong:
10 and 15 are too small and arise if we only consider one group size pattern or incorrectly treat boxes as labelled or partly labelled.
20 is closer but still undercounts, often because some patterns of group sizes are omitted or double counting is removed incorrectly.

Common Pitfalls:
Many students mistakenly treat the boxes as distinct and simply distribute toys into three labelled boxes, then try to adjust for identical boxes in an ad hoc way. Others forget the nonempty condition and include partitions with empty boxes. For identical boxes with a fixed number of nonempty boxes, always think in terms of partitions of a set into unlabeled subsets and, when possible, use Stirling numbers of the second kind.

Final Answer:
The number of ways to pack the toys is 25.