Counting Without Double Counting

Combinatorics is the art of counting things without actually listing all of them.

That sounds simple until the same object gets counted twice, a case gets missed, or the problem looks like it has too many possibilities to manage.

The real skill is not memorizing formulas. It is learning how to organize the world.

The First Rule: Define What You Are Counting

Before doing any math, ask:

What exactly counts as one object?

For example, if you are arranging the letters in MATH, then MATH and TAMH are different arrangements.

But if you are choosing a team of three students, the order does not matter. Choosing Ava, Ben, and Chris is the same team as choosing Chris, Ava, and Ben.

This distinction is the difference between permutations and combinations.

Cases: Split the Problem Cleanly

Cases are useful when the problem naturally breaks into separate situations.

Example: How many two-digit numbers have at least one digit equal to 7?

You could split into:

  • Tens digit is 7
  • Ones digit is 7

But that double-counts 77.

A cleaner split is:

  • Tens digit is 7: 10 numbers
  • Tens digit is not 7, but ones digit is 7: 8 numbers

Total: 18 numbers.

The key is that cases should not overlap unless you plan to subtract the overlap.

Complements: Count What You Do Not Want

Sometimes it is easier to count the opposite.

Example: How many three-digit numbers have at least one digit equal to 5?

There are 900 three-digit numbers total.

Now count the ones with no 5:

  • Hundreds digit: 8 choices, because it can be 1–9 except 5
  • Tens digit: 9 choices, because it can be 0–9 except 5
  • Ones digit: 9 choices

So there are 8⋅9⋅9=6488 \cdot 9 \cdot 9 = 648 numbers with no 5.

Therefore, the answer is 900−648=252900 - 648 = 252.

Symmetry: Let the Problem Do Work for You

If all outcomes are evenly balanced, symmetry can save huge amounts of effort.

Example: In all arrangements of the letters A, B, C, D, how many have A before B?

Exactly half.

Why? For every arrangement where A comes before B, swap A and B to get one where B comes before A. This pairing is perfect, so the answer is 4!2=12\frac{4!}{2} = 12.

This is a bijection: a one-to-one matching between two sets.

COSMO Challenge

How many ways can you arrange the letters in COSMO?

Be careful: the letter O appears twice.

First count as if all five letters were different: 5!=1205! = 120.

But the two O’s can be swapped without changing the arrangement, so every real arrangement has been counted twice.

Final answer: 5!2!=60\frac{5!}{2!} = 60.

Now try MISSISSIPPI and explain what you divide by.

Takeaway

Good counting is not speed. It is control.

When students learn to define objects, split cases cleanly, count complements, and use symmetry, they start to see combinatorics as a way of thinking, not a bag of formulas.

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