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Probability and StatisticsGrades 9–123 min read

Fundamental Counting Principle

The fundamental counting principle multiplies the number of choices across successive stages to count complete outcomes efficiently.

Cheat sheet
If a process has several stages, multiply the number of available choices at each stage—after accounting for any restrictions.

The multiplication principle

If stage one can occur in $m$ ways and, for each of those, stage two can occur in $n$ ways, then the two-stage process has

$$ mn $$

possible outcomes. With more stages, continue multiplying the number of choices available at each stage.

Why multiplication works

Imagine a tree diagram. Each of $m$ first branches produces $n$ second branches, giving $n$ outcomes repeated across $m$ groups. The total is repeated addition:

$$ \underbrace{n+n+\cdots+n}_{m\text{ groups}}=mn. $$

The principle compresses a large tree without listing every leaf.

Worked example: independent stages

The stages need not be probabilistically independent; what matters is knowing the available count at each branch.

When choices depend on earlier choices

Suppose a three-letter code uses distinct letters from an alphabet of $26$. The first position has $26$ choices, the second $25$, and the third $24$:

$$ 26\cdot25\cdot24. $$

The choice counts change because repetition is forbidden. Build stages in the order that makes restrictions easy to track.

Addition versus multiplication

Use multiplication for stages joined by and: choose a main dish and a drink. Use addition for disjoint alternatives joined by or: choose one item from the breakfast menu or one from the lunch menu.

If alternatives overlap, subtract the overlap or split them into genuinely disjoint cases before adding.

Casework

Some restrictions create branches with different counts. Count each disjoint case using multiplication, then add the case totals.

For example, a code may begin with either a letter followed by digits or a digit followed by letters. Count each format separately because their stage structures differ.

Complement counting

It is often easier to count all outcomes and subtract unwanted outcomes:

$$ N(\text{desired})=N(\text{total})-N(\text{undesired}). $$

“At least one” problems often benefit from subtracting the case with none.

Repetition rules

Clarify whether choices may repeat. A four-digit PIN with repetition allowed has $10^4$ possibilities. If all digits must differ, the count is

$$ 10\cdot9\cdot8\cdot7. $$

If the first digit cannot be zero, its stage has only $9$ choices, while later stages may have different counts.

Connecting to permutations and combinations

Permutation formulas are multiplication-principle shortcuts for ordered selections without replacement. Combination formulas remove the repeated orderings when order does not matter.

Understanding stage counting makes the formulas easier to choose and adapt.

Probability applications

For equally likely outcomes,

$$ P(A)=\frac{N(A)}{N(S)}. $$

The counting principle can determine both favourable and total outcome counts. Verify that the sample space outcomes are genuinely equally likely before using the ratio.

Common mistakes

Adding stage counts instead of multiplying. A complete outcome needs one choice from every stage.

Using the same count after a no-repetition choice. Available options decrease.

Treating overlapping cases as disjoint. This double-counts shared outcomes.

Forgetting a leading-zero restriction. Codes and numbers may have different rules.

Assuming order never matters. Define what makes two outcomes distinct.

Quick self-check

  • What are the stages of one complete outcome?
  • How many choices are available at each stage after earlier decisions?
  • Is repetition allowed?
  • Are alternatives disjoint before their counts are added?
  • Would complement counting be simpler?
  • Does the answer distinguish ordered and unordered outcomes correctly?
Check your understanding

Try it yourself

Hints are part of learning. Open one whenever it makes the next step feel possible.

1 practice question
Question 1Count a multi-stage outcome · Gentle

A student chooses 1 of 4 shirts, 1 of 3 pairs of pants, and 1 of 2 pairs of shoes. How many outfits are possible?

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