In probability you counted permutations by multiplying choices. That multiplication is one of two basic rules of counting. Both rules are facts about sets you have seen: the size of a union of separate sets, and the size of a Cartesian product.
Idea clave
Sum rule: if a choice can be made in one of several ways that cannot happen together, with $a$, $b$, … options in each, the total is $a + b + \cdots$. Product rule: if a choice is made in steps, with $a$ options for the first step and $b$ for the second (whatever the first was), and so on, the total is $a \cdot b \cdots$.
Por qué funciona
The sum rule is $|A \cup B| = |A| + |B|$ for disjoint sets: nothing is counted twice because nothing is in both. The product rule is $|A \times B| = |A| \cdot |B|$: a two-step choice is an ordered pair, and each first step pairs with every second step. The words to listen for: "either … or" (alternatives, add) versus "first … then" (a sequence, multiply). Big problems often need both: split into cases with the sum rule, count each case with the product rule.
Worked examples
Ejemplo resuelto: A license plate
A plate has $3$ letters (A–Z) followed by $3$ digits (0–9), repeats allowed. How many plates are there?
- Six steps in a row: $26$ choices for each letter and $10$ for each digit.
- Product rule: $26^{3} \cdot 10^{3} = 17{,}576 \cdot 1{,}000 = 17{,}576{,}000$.
Respuesta: $17{,}576{,}000$
Ejemplo resuelto: Two possible lengths
A code uses digits 0–9 and has length $3$ or $4$. How many codes are there?
- A code has one length or the other, never both, so split into cases and add.
- Length $3$: $10^{3} = 1{,}000$. Length $4$: $10^{4} = 10{,}000$.
- Sum rule: $1{,}000 + 10{,}000 = 11{,}000$.
Respuesta: $11{,}000$
Ejemplo resuelto: Cases, then products
How many three-digit numbers are even and have three different digits?
- The last digit must be even, and $0$ behaves differently because the first digit cannot be $0$. So split by the last digit.
- Last digit $0$: first digit $1$–$9$ ($9$ ways), middle digit anything else ($8$ ways): $9 \cdot 8 = 72$.
- Last digit $2$, $4$, $6$ or $8$ ($4$ ways): first digit not $0$ and not the last digit ($8$ ways), middle digit not either of those ($8$ ways): $4 \cdot 8 \cdot 8 = 256$.
- Sum rule: $72 + 256 = 328$.
Respuesta: $328$
Error común
Multiplying alternatives or adding a sequence. "Choose a dessert: one of $4$ pies or one of $3$ cakes" is $4 + 3 = 7$, not $12$; there is no second step. "Choose a pie and a cake" is $4 \cdot 3 = 12$, not $7$. Check: ask whether one outcome involves both choices (multiply) or just one of them (add).