Multiplication Rule for Counting¶

Definition¶

Consider a compound experiment made of two stages. If the first stage has $a$ possible outcomes and, for each of them, the second stage has $b$ possible outcomes, then the compound experiment has

a\cdot b

possible outcomes. This is the multiplication rule for counting. For $k$ successive stages with $n_1,n_2,\ldots,n_k$ choices it gives $n_1n_2\cdots n_k$ ordered outcomes; in set notation, $|A\times B|=|A|\,|B|$ $∣ A ∣ ∣ B ∣$ .

Interpretation¶

The multiplication rule counts ordered outcomes stage by stage: in a tree diagram, the number of leaves is the product of the numbers of branches at each level. The same tree carries probabilities, and multiplying along a path gives the probability of that path.

A tree for drawing two cards: ace or not ace on the first draw, then ace or not ace on the second, with conditional probabilities on each branch; two aces have probability 1/221.

Counting the ordered outcomes gives $52\cdot51=2652$ ordered draws, $4\cdot3=12$ of them two aces, so $\tfrac{12}{2652}=\tfrac{1}{221}$ $\frac{1}{221}$ . Probabilities multiply along a path in the same way: $\mathbb{P}(A\cap B)=\mathbb{P}(A)\,\mathbb{P}(B\mid A)$ $A)$ .

The Cartesian product records an ordered choice from each stage. If every first-stage choice permits the same $b$ second-stage choices, counting the pairs gives $|A\times B|=|A||B|$ $∣ A ∣∣ B ∣$ .

Key Results¶

The number of choices at a stage must not depend on the outcomes of earlier stages, although the choices themselves may.
Probability version: for events $A$ and $B$ with $\mathbb{P}(A)>0$ $0$ , $\mathbb{P}(A\cap B)=\mathbb{P}(A)\mathbb{P}(B\mid A)$ $A)$ ; for several events the rule becomes a product of successive conditional probabilities.
For independent events, the conditional factor reduces to the marginal probability.

Examples¶

A code of two letters followed by three digits can be formed in $26^2\cdot10^3=676\,000$ ways.
Two dice give $6\cdot6=36$ ordered outcomes, and $n$ distinct objects can be ordered in $n(n-1)\cdots1=n!$ ways.
Two cards drawn without replacement are both aces with probability $\frac{4}{52}\cdot\frac{3}{51}=\frac{1}{221}$ $\frac{1}{221}$ : the chance of the first ace times the conditional chance of the second. In counts, $4\cdot3=12$ of the $52\cdot51=2652$ ordered draws.

Prerequisites¶

Sources¶

Probability Theory for EOR · Week 1

Aliases: rule of product, fundamental counting principle, multiplication principle, basic principle of counting Keywords: product rule for counting Symbols: intersection

Last updated: 17 October 2026 21:53