Counting Principle

The counting principle counts complete choices without listing every one. It is the foundation of arrangements, selections and many probability calculations.

Two ideas do most of the work: multiply across successive stages when the number of choices at each stage is known, and add across disjoint alternatives. A count by itself is not yet a probability; the selection model must still tell us whether the counted outcomes are equally likely.

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Multiply across successive choices

Suppose a workshop activity is specified by a project, a time slot and a kit. There are three projects, two slots available for each project, and two kits available for every project-slot pair. The number of complete selections is

3\times2\times2=12.

Each complete choice consists of one option at each stage. A systematic list could be built by writing four slot-kit combinations under each project; multiplication compresses that list.

Count complete combinations of staged choices

Every project permits both slots and both kits. These are choice counts, not assumed probabilities.

Diagram explanation and exact data

Each of three rows contains four slot-kit choices. If one project had fewer available slots or kits, the rectangular product model would need adjustment.

Count complete combinations of staged choices: exact reference data
Project	Slot	Kit
P1	1	A
P1	1	B
P1	2	A
P1	2	B
P2	1	A
P2	1	B
P2	2	A
P2	2	B
P3	1	A
P3	1	B
P3	2	A
P3	2	B

In general, if there are $m_1$ first-stage choices, $m_2$ second-stage choices after each first-stage choice, and so on, the count is $m_1m_2\cdots m_r$ . The number at a stage may depend on the stage, but a simple product requires that it be the same for every history reaching that stage.

This is a counting rule, not a statistical-independence statement. Selecting different people for successive roles is dependent when sampling is involved, yet the ordered selections can still be counted by a product of shrinking choice counts.

Add across disjoint alternatives

Suppose an activity is either one of three studio sessions or one of two field sessions, with no session counted in both categories. There are $3+2=5$ choices. The word “either” describes alternatives, not successive decisions, so multiplying would count the wrong object.

If categories overlap, subtract the overlap or split them into disjoint cases first. For example, eight workshops include art and six include technology, with three including both. There are $8+6-3=11$ distinct workshops in at least one category.

This is the counting counterpart of the Addition Rule of Probability.

Handle branches with different numbers of options

Suppose project A has three available sessions, but project B has two. The number of project-session choices is $3+2=5$ . There is no single second-stage count of 3 that applies to both projects, so $2\times3$ would overcount.

For more complex constraints, break the problem into disjoint cases where a simple product does apply, calculate each case, then add. A small tree is often the clearest way to see which histories have different branch counts.

Codes and numbers have different leading-zero rules

A four-position code may allow any digit 0–9 in every position. With repetition allowed, there are $10^4=10{,}000$ codes. If digits cannot repeat, the count becomes

10\times9\times8\times7=5{,}040.

A four-digit positive integer has a different rule: its first digit cannot be zero. With no repeated digits, the count is

9\times9\times8\times7=4{,}536.

There are nine first choices. After one non-zero digit is used, nine digits remain for the second position, including zero. Subtracting zero from every position would be another error: it is forbidden only at the beginning.

State whether the object is a code, a written integer, an arrangement of distinct objects or a selection. Those words determine the counting conditions.

Count a complement when the restriction is awkward

How many four-position digit codes contain at least one zero if repetition is allowed? Count all codes and subtract codes with no zero:

10^4-9^4=10{,}000-6{,}561=3{,}439.

Counting the possible position of “a zero” directly would count codes with several zeros more than once. Complementing the no-zero case avoids that overlap.

The corresponding probability is $3{,}439/10{,}000$ only if the code is chosen uniformly from all 10,000 codes. Human-selected codes do not automatically satisfy that model.

Factorials describe successive choices without replacement

Arranging five distinct books on a shelf gives five first choices, four second choices, and so on:

5!=5\times4\times3\times2\times1=120.

The factorial notation $n!$ records that product for a positive integer n. The convention $0!=1$ corresponds to one empty arrangement and keeps later formulas consistent.

If only three ordered positions are filled from five distinct books, the count stops after three factors: $5\times4\times3=60$ . Permutations develops ordered arrangements, while Combinations removes the order when only membership matters.

Check what one counted outcome means

Before dividing favourable by total counts, confirm that both counts describe the same kind of object. Ordered token draws should be compared with ordered draws, and unordered subsets with unordered subsets. Mixing the two can introduce an extra factorial or omit one.

Also check that each outcome is counted exactly once. Repeated symbols, equivalent rotations and constraints can make apparently different descriptions represent the same result. The counting formulas in later lessons are useful because they handle those equivalences explicitly, not because every problem automatically fits a formula.

Practice questions

Four routes each offer three departure times. How many route-time choices are there?
How many four-digit positive integers have no repeated digit?
How many four-position digit codes contain at least one zero when repetition is allowed?

Check your answers

$4\times3=12$ , provided all three times are available for each route.
$9\times9\times8\times7=4{,}536$ . Zero is excluded only from the first position.
Subtract no-zero codes from all codes: $10^4-9^4=3{,}439$ .

Sources and further reading

OpenStax: Counting principles.