A Fairly Simple Math Problem from My Middle School Years — Still Useful Today

Created on June 05, 2025

2025

For reasons both universally acknowledged and personally unclear, I’ve always had a quiet fondness for math class in middle school. Recently, while working through a research problem, I ran into a version of a Permutation and Combination question that felt straight out of those school days. And it turned out to be surprisingly useful — not just for logic puzzles, but also for counting design combinations, modeling probabilities, and even generating constrained inputs in machine learning setups.

The Problem (a simplified version, unrelated to the High Entropy Alloy work I’m actually doing)

You have 10 drawers arranged in sequence, and each drawer can hold one of five different colored balls (say: A, B, C, D, E).

The constraint: All five colors must appear at least once.

  • Question 1: How many valid arrangements are there?
  • Question 2: Given that all five appear, if you pick one color at random, what’s the probability that it appears exactly 1, 2, … or 6 times?
  • Question 3: Could these probabilities all be made equal by design? (Spoiler: No. We’ll see why.)

Question 1 — Total Valid Combinations

If we ignore the constraint, each drawer can independently take on one of five colors:

\[5^{10} = 9,\!765,\!625\]

But we want only the combinations where all five colors appear at least once. This is where the Inclusion–Exclusion Principle comes in.

Inclusion–Exclusion: What It Does

The Inclusion–Exclusion Principle helps calculate the size of a union of overlapping sets by correcting for overcounts.

Let’s say you have sets (A_1, A_2, \dots, A_n), and you want to count how many elements are in at least one of them:

\[|A_1 \cup A_2 \cup \dots \cup A_n| = \sum |A_i| - \sum |A_i \cap A_j| + \sum |A_i \cap A_j \cap A_k| - \dots + (-1)^{n+1}|A_1 \cap \dots \cap A_n|\]

In our case, we’re not looking for the union — we want to count how many assignments include all 5 colors, which is the complement of the union of the assignments excluding at least one color.

That gives us:

\[T = \sum_{i=0}^{5} (-1)^i \binom{5}{i} (5 - i)^{10}\]

Where:

  • (\binom{5}{i}) = ways to choose i colors to exclude
  • ((5 - i)^{10}) = number of 10-drawer sequences using only the remaining colors
  • ((-1)^i) = alternates between subtracting and adding to adjust for overlapping exclusions

The binomial coefficient (\binom{5}{i}) is calculated as:

\[\binom{5}{i} = \frac{5!}{i!(5 - i)!}\]

Running this gives us:

5,103,000 valid designs where all five colors are used at least once.

Question 2 — Probability That a Given Color Appears k Times

Let’s say we fix one color, like “A.” What’s the probability that it appears exactly k times, for (1 \leq k \leq 6), in a valid design?

Strategy

  1. Choose (k) positions out of 10 to assign to “A”: (\binom{10}{k})
  2. Assign the remaining (10 - k) positions to the other 4 colors, ensuring each appears at least once

We use inclusion–exclusion again for this:

\[P(K = k) = \frac{\binom{10}{k} \sum_{j=0}^{4} (-1)^j \binom{4}{j} (4 - j)^{10 - k}}{T}\]

Where:

  • (T) is the total valid designs from Question 1
  • The numerator counts how many ways to place “A” exactly (k) times, and the rest validly among the other four

Results

Each of these probabilities corresponds to a fraction of the form:

\[P(K = k) = \frac{\text{Number of valid assignments where A appears } k \text{ times}}{T}\]

Using that, we get:

  • P(appears 1 time) = 0.365432 = 1,865,481 / 5,103,000
  • P(appears 2 time) = 0.360000 = 1,837,080 / 5,103,000
  • P(appears 3 times) = 0.197531 = 1,008,547 / 5,103,000
  • P(appears 4 times) = 0.064198 = 327,607 / 5,103,000
  • P(appears 5 times) = 0.011852 = 60,514 / 5,103,000
  • P(appears 6 times) = 0.000988 = 5,771 / 5,103,000

Question 3 — Can These Probabilities Be Made Equal?

Suppose we assume that all probabilities are equal and denote this common value by (P). That is,

\[P(K = 1) = P(K = 2) = \dots = P(K = 6) = P\]

Now let’s say we generate 1,000 valid samples. If each value of (K = k) appears (d) times, then:

\[6 \cdot d = 1000 \quad \Rightarrow \quad d = \frac{1000}{6}\]

This would also imply the total number of times color A appears across all samples is:

\[1 \cdot d + 2 \cdot d + 3 \cdot d + 4 \cdot d + 5 \cdot d + 6 \cdot d = (1+2+3+4+5+6) \cdot d = 21d\]

However, since there are 10,000 layer positions (1,000 samples × 10 layers) and the five colors must appear equally often, color A must occupy exactly:

\[\frac{10,000}{5} = 2000 \text{ positions}\]

So we would need:

\[21d = 2000 \quad \Rightarrow \quad d = \frac{2000}{21}\]

But earlier we said (d = \frac{1000}{6}). These two values of (d) are not equal:

\[\frac{1000}{6} \ne \frac{2000}{21}\]

This contradiction shows that our assumption — that all probabilities (P(K = k)) are equal — cannot hold under the constraint that each color appears equally across the dataset.

Final Thoughts

At first, I didn’t give Question 3 much thought. I assumed it might be possible to make the probabilities equal, so I jumped straight into trying to write code for it. But halfway through, I realized it wasn’t working — not because of a bug, but because the logic didn’t hold. That’s when I stepped back and worked through the math.

Lesson learned: think it through before jumping into code — especially when the constraints are doing more than they seem on the surface.