Permutations and Combinations • Topic 1 of 3

Fundamental Principle of Counting and Factorial

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.

$$\text{ways} = m_1 \times m_2 \times m_3 \times \cdots \times m_k$$

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.

SituationSignal wordPrincipleOperation
Do step 1, then step 2and / thenMultiplication$m \times n$
Pick option A or option BorAddition$m + n$

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 $10! = 3{,}628{,}800$. Two facts must be memorised. First, the recursive relation $n! = n \times (n-1)!$, which lets you simplify ratios of factorials without ever expanding them fully. 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 $0$.

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.

Multiplication principle filling slots for a meal choice Multiplication Principle: 3 starters AND 4 mains STARTER3 choices × MAIN4 choices = 12 meals M1M2M3M4 each starter branches into 4 mains → 3 × 4 = 12 Filling four ordered boxes from four distinct objects gives four factorial Arranging 4 objects: 4! = 24 4 3 2 1 ××× choices shrink by 1 each box: 4 × 3 × 2 × 1 = 24
1
Worked Example
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
  1. The choices are made in sequence: pick a sandwich and a drink and a dessert — the word "and" signals the multiplication principle.
  2. Sandwich: $3$ ways. Drink: $4$ ways. Dessert: $2$ ways.
  3. Total $= 3 \times 4 \times 2 = 24$.

Answer: $24$ different combo meals.

2
Worked Example
How many three-digit numbers can be formed using the digits $1, 2, 3, 4, 5$ if no digit is repeated?
Solution
  1. There are three ordered positions to fill: hundreds, tens, units.
  2. Hundreds place: any of $5$ digits.
  3. Tens place: any of the remaining $4$ digits (one is used up).
  4. Units place: any of the remaining $3$ digits.
  5. Total $= 5 \times 4 \times 3 = 60$.

Answer: $60$ three-digit numbers.

3
Worked Example
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
  1. The student picks a science elective or a humanities elective — the alternatives are mutually exclusive, so use the addition principle.
  2. Science: $5$ ways. Humanities: $3$ ways.
  3. Total $= 5 + 3 = 8$.

Answer: $8$ ways.

4
Worked Example
Evaluate $\dfrac{8!}{5!}$ without fully expanding either factorial.
Solution
  1. Use $8! = 8 \times 7 \times 6 \times 5!$, so $\dfrac{8!}{5!} = \dfrac{8 \times 7 \times 6 \times 5!}{5!}$.
  2. The $5!$ cancels: $\dfrac{8!}{5!} = 8 \times 7 \times 6$.
  3. $8 \times 7 = 56$, and $56 \times 6 = 336$.

Answer: $\dfrac{8!}{5!} = 336$.

5
Worked Example
Find the value of $n$ if $(n+1)! = 12 \times (n-1)!$.
Solution
  1. Write $(n+1)! = (n+1) \times n \times (n-1)!$.
  2. Substitute: $(n+1) \times n \times (n-1)! = 12 \times (n-1)!$.
  3. Cancel $(n-1)!$ from both sides: $(n+1)\,n = 12$, i.e. $n^2 + n - 12 = 0$.
  4. Factorise: $(n+4)(n-3) = 0$, giving $n = 3$ or $n = -4$.
  5. A factorial needs a non-negative integer, so reject $n = -4$.

Answer: $n = 3$.

6
Worked Example
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
  1. Thousands place cannot be $0$, so it has $5$ choices (any of $1,2,3,4,5$).
  2. Hundreds place: $0$ is now allowed again, but one non-zero digit is used — $5$ choices remain ($0$ plus the four unused).
  3. Tens place: $4$ choices remain. Units place: $3$ choices remain.
  4. Total $= 5 \times 5 \times 4 \times 3 = 300$.

Answer: $300$ four-digit numbers.

Key Points

  • Multiplication principle: choices in sequence ("and / then") multiply — $m_1 \times m_2 \times \cdots$.
  • Addition principle: mutually exclusive alternatives ("or") add — $m + n$.
  • Factorial: $n! = n \times (n-1) \times \cdots \times 1$, with the recursion $n! = n \times (n-1)!$.
  • By convention $0! = 1$ — forced by the recursion and the single empty arrangement.
  • Simplify factorial ratios by cancelling, e.g. $\dfrac{8!}{5!} = 8 \times 7 \times 6$ — never expand fully.
Tap an option to check your answer0 / 4
Q1.$5!=$
Explanation: $5!=5\cdot4\cdot3\cdot2\cdot1=120$.
Q2.By the fundamental principle of counting, if one task can be done in $m$ ways and another in $n$ ways, both can be done in:
Explanation: Multiply the number of ways.
Q3.$0!=$
Explanation: By convention $0!=1$.
Q4.The number of ways to fill $2$ distinct posts from $4$ people is:
Explanation: $4\times3=12$.