Title: How Many Ways to Distribute 4 Distinct Chemical Samples into 2 Identical Containers with At Least One Sample Each?

When working with discrete objects like chemical samples and containers with unique constraints, combinatorics becomes both essential and fascinating. One common yet cleverly non-trivial problem is: How many ways can you distribute 4 distinct chemical samples into 2 identical storage containers, ensuring that no container is empty?

This problem lies at the intersection of combinatorics and logistics—particularly relevant in laboratories, supply chains, and quality control scenarios. Let’s unpack the solution step-by-step to uncover how many valid distributions satisfy the condition that each container holds at least one sample, and the containers themselves cannot be told apart.

Understanding the Context

Understanding the Constraints

  • The samples are distinct: Sample A, B, C, and D are unique.
  • The containers are identical: Placing samples {A,B} in Container 1 and {C,D} in Container 2 is the same distribution as the reverse.
  • Each container must contain at least one sample — no empties allowed.
  • We seek distinct distributions up to container symmetry.
$$Question: How many ways are there to distribute 4 distinct chemical samples into 2 identical storage containers such that each container has at least one sample?

Key Insights

Step 1: Count Total Distributions Without Identical Containers

If the containers were distinguishable (e.g., “Container X” and “Container Y”), distributing 4 distinct samples into 2 labeled containers results in:

> $ 2^4 = 16 $ possible assignments (each sample independently assigned to one of the two containers).

However, we must exclude the 2 cases where all samples go to one container:

  • All in Container 1
  • All in Container 2

So total distributions with non-empty containers (distinguishable containers):
$$
16 - 2 = 14
$$

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Final Thoughts

Step 2: Adjust for Identical Containers

When containers are identical, distributions that differ only by swapping containers are considered the same. For example:

  • {A,B} | {C,D} ↔ {C,D} | {A,B} — same configuration.

To count distinct distributions with identical containers and non-empty subsets, we must group these identical partitions.

This is a classic combinatorics problem solved by considering partitions of a set.

Using Set Partitions: Stirling Numbers of the Second Kind

The number of ways to partition a set of $ n $ distinct objects into $ k $ non-empty, unlabeled subsets is given by the Stirling number of the second kind, denoted $ S(n, k) $.

For our case:

  • $ n = 4 $ chemical samples
  • $ k = 2 $ containers (non-empty, identical)

We compute:
$$
S(4, 2) = 7
$$