Counting sounds like the most elementary thing in mathematics, yet beyond a handful of objects, listing every possibility by hand becomes hopeless. The whole of this chapter exists to count without listing. Two principles do the heavy lifting, and almost every counting question reduces to deciding which one applies.
Multiplication Principle (Fundamental Principle of Counting, FPC): If one task can be done in $m$ ways and, after it is done, a second independent task can be done in $n$ ways, then the two tasks together can be done in $m \times n$ ways. The phrase that signals multiplication is "and then" — a sequence of choices made one after another. The principle extends to any number of stages.
$$\text{ways} = m_1 \times m_2 \times m_3 \times \cdots \times m_k$$
Addition Principle: If a task can be done by method A in $m$ ways or by method B in $n$ ways, and the two methods cannot happen together, then the task can be done in $m + n$ ways. The signal word here is "or" — a choice between mutually exclusive alternatives. The "mutually exclusive" condition is essential: if the alternatives can overlap, the overlap must be subtracted to avoid double-counting.
Factorial. Counting arrangements forces us to multiply long descending strings of integers, so we give that product a name. For a positive integer $n$, $n$ factorial is the product of all positive integers up to $n$:
$$n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1$$
Factorials grow ferociously fast: $5! = 120$ but $10! = 3{,}628{,}800$. Several facts are worth memorising. First, the recursive relation $n! = n \times (n-1)!$, which lets you simplify ratios of factorials without ever expanding them fully — for example $\dfrac{n!}{(n-1)!} = n$ and $\dfrac{(n+2)!}{n!} = (n+2)(n+1)$. Second, by convention:
$$0! = 1$$
Why is $0! = 1$? It is not a fudge. The recursion $n! = n \times (n-1)!$ at $n = 1$ gives $1! = 1 \times 0!$, which forces $0! = 1$. There is also a combinatorial reason: there is exactly one way to arrange nothing — the empty arrangement — so the count of arrangements of zero objects must be $1$, not $0$.
Simplifying factorial expressions. The single most useful manoeuvre is to peel a larger factorial down to a smaller one and cancel. Whenever you meet a ratio such as $\dfrac{n!}{(n-r)!}$, expand only the top until the bottom factorial appears, then cancel it whole. Note also that $n! + (n+1)! = n!\,[1 + (n+1)] = (n+2)\,n!$, and that factorials are never distributive over addition — $(a+b)! \ne a! + b!$.
Deeper Insight — one decision tree, two operations: Beginners treat the multiplication and addition principles as separate tricks to memorise, but they are really two readings of the same picture: a decision tree. When choices happen in sequence — first this, then that — the branches multiply, because each early branch splays into a full fan of later ones. When choices are alternatives — this route or that route — the branches add, because you travel down only one of them. The single most reliable habit in counting is to narrate the problem to yourself in plain words and listen for "and" versus "or"; the arithmetic then writes itself. Factorial is simply what the multiplication principle produces when you arrange all of a set's objects in a row — $n$ choices for the first slot, $n-1$ for the next, and so on down to $1$ — which is why it sits at the root of both permutations and combinations.
A canteen offers 3 kinds of sandwiches, 4 kinds of drinks, and 2 kinds of desserts. In how many ways can a student choose one of each to make a combo meal?
Solution- The choices are made in sequence: pick a sandwich and a drink and a dessert — the word "and" signals the multiplication principle.
- Sandwich: $3$ ways. Drink: $4$ ways. Dessert: $2$ ways.
- Total $= 3 \times 4 \times 2 = 24$.
Answer: $24$ different combo meals.
How many three-digit numbers can be formed using the digits $1, 2, 3, 4, 5$ if no digit is repeated?
Solution- There are three ordered positions to fill: hundreds, tens, units.
- Hundreds place: any of $5$ digits.
- Tens place: any of the remaining $4$ digits (one is used up).
- Units place: any of the remaining $3$ digits.
- Total $= 5 \times 4 \times 3 = 60$.
Answer: $60$ three-digit numbers.
A student must choose a single elective: either one of 5 science electives or one of 3 humanities electives. In how many ways can the choice be made?
Solution- The student picks a science elective or a humanities elective — the alternatives are mutually exclusive, so use the addition principle.
- Science: $5$ ways. Humanities: $3$ ways.
- Total $= 5 + 3 = 8$.
Answer: $8$ ways.
Evaluate $\dfrac{8!}{5!}$ without fully expanding either factorial.
Solution- Use $8! = 8 \times 7 \times 6 \times 5!$, so $\dfrac{8!}{5!} = \dfrac{8 \times 7 \times 6 \times 5!}{5!}$.
- The $5!$ cancels: $\dfrac{8!}{5!} = 8 \times 7 \times 6$.
- $8 \times 7 = 56$, and $56 \times 6 = 336$.
Answer: $\dfrac{8!}{5!} = 336$.
Find the value of $n$ if $(n+1)! = 12 \times (n-1)!$.
Solution- Write $(n+1)! = (n+1) \times n \times (n-1)!$.
- Substitute: $(n+1) \times n \times (n-1)! = 12 \times (n-1)!$.
- Cancel $(n-1)!$ from both sides: $(n+1)\,n = 12$, i.e. $n^2 + n - 12 = 0$.
- Factorise: $(n+4)(n-3) = 0$, giving $n = 3$ or $n = -4$.
- A factorial needs a non-negative integer, so reject $n = -4$.
Answer: $n = 3$.
How many four-digit numbers can be formed using the digits $0, 1, 2, 3, 4, 5$ if repetition is not allowed? (A four-digit number cannot start with $0$.)
Solution- Thousands place cannot be $0$, so it has $5$ choices (any of $1,2,3,4,5$).
- Hundreds place: $0$ is now allowed again, but one non-zero digit is used — $5$ choices remain ($0$ plus the four unused).
- Tens place: $4$ choices remain. Units place: $3$ choices remain.
- Total $= 5 \times 5 \times 4 \times 3 = 300$.
Answer: $300$ four-digit numbers.
Compute $\dfrac{1}{6!} + \dfrac{1}{7!} = \dfrac{x}{8!}$ and find $x$.
Solution- Write every term over the common denominator $8!$. Since $8! = 8 \times 7 \times 6!$ and $8! = 8 \times 7!$:
- $\dfrac{1}{6!} = \dfrac{8 \times 7}{8!} = \dfrac{56}{8!}$ and $\dfrac{1}{7!} = \dfrac{8}{8!}$.
- Add: $\dfrac{56}{8!} + \dfrac{8}{8!} = \dfrac{64}{8!}$.
Answer: $x = 64$.
Simplify $\dfrac{(n+2)! + (n+1)!}{(n+1)!}$ in terms of $n$.
Solution- Factor the numerator: $(n+2)! = (n+2)(n+1)!$, so $(n+2)! + (n+1)! = (n+1)!\,[(n+2) + 1]$.
- That gives $(n+1)!\,(n+3)$.
- Divide by $(n+1)!$: the factorial cancels, leaving $n+3$.
Answer: $\dfrac{(n+2)! + (n+1)!}{(n+1)!} = n + 3$.
A car number plate has the format: two letters (A–Z) followed by four digits (0–9), and repetition is allowed throughout. How many distinct plates are possible?
Solution- Each position is filled independently, so use the multiplication principle across all six positions.
- Two letter positions: $26 \times 26 = 26^2$ ways. Four digit positions: $10 \times 10 \times 10 \times 10 = 10^4$ ways.
- Total $= 26^2 \times 10^4 = 676 \times 10{,}000$.
Answer: $6{,}760{,}000$ distinct plates.
How many three-digit numbers between $100$ and $999$ are even and use distinct digits? (Digits $0$–$9$, no repetition.)
Solution- A three-digit number is even when its units digit is $0, 2, 4, 6,$ or $8$. Split into two cases, because choosing $0$ for the units frees up the hundreds digit differently.
- Case 1: units digit $= 0$. Hundreds: $9$ choices ($1$–$9$); tens: $8$ remaining choices. That is $9 \times 8 = 72$.
- Case 2: units digit $\in \{2,4,6,8\}$ ($4$ choices). Hundreds cannot be $0$ or the used units digit, so $8$ choices; tens: any of the $8$ remaining digits. That is $4 \times 8 \times 8 = 256$.
- Add the mutually exclusive cases: $72 + 256 = 328$.
Answer: $328$ such numbers.
In how many ways can $5$ different prizes be awarded to $3$ students if each student is eligible for every prize and may receive any number of prizes?
Solution- Think prize-by-prize: each of the $5$ prizes is handed to one of the $3$ students — a sequence of $5$ independent choices.
- Each prize has $3$ recipients to choose from, so by the multiplication principle the count is $3 \times 3 \times 3 \times 3 \times 3 = 3^5$.
- $3^5 = 243$.
Answer: $243$ ways.