The Fundamental Counting Principle#
If one event can happen in $m$ ways, and a second event can happen in $n$ ways, then both events together can happen in $m \times n$ ways.
Why Multiplication?#
Think of choosing an outfit: 3 shirts and 4 pants.
For each of the 3 shirts, you can pair it with any of the 4 pants. That’s $3 + 3 + 3 + 3 = 3 \times 4 = 12$ combinations.
This extends to any number of events: if there are $k$ events with $n_1, n_2, \dots, n_k$ options respectively, the total number of outcomes is:
$$n_1 \times n_2 \times n_3 \times \dots \times n_k$$Key insight: We multiply because each choice is independent — each option at one stage can combine with every option at every other stage.
1. The Slot Method#
For arrangement problems, draw one slot for each position. Write the number of available choices above each slot, then multiply.
Worked Example 1 — Codes with Repetition#
$$\underset{\text{1st letter}}{[\;5\;]} \times \underset{\text{2nd letter}}{[\;5\;]} \times \underset{\text{3rd letter}}{[\;5\;]} = 125$$How many 3-letter codes can be made from $\{A, B, C, D, E\}$ if letters may be repeated?
Each slot has 5 options because repetition is allowed — after using a letter, it’s still available.
Worked Example 2 — Codes without Repetition#
$$\underset{\text{1st}}{[\;5\;]} \times \underset{\text{2nd}}{[\;4\;]} \times \underset{\text{3rd}}{[\;3\;]} = 60$$Same letters, but no repetition allowed.
After choosing the 1st letter (5 options), only 4 remain for the 2nd, then 3 for the 3rd.
Worked Example 3 — PIN Codes#
$$10 \times 10 \times 10 \times 10 = 10^4 = 10\,000$$How many 4-digit PINs are possible using digits 0–9 (repetition allowed)?
$$10 \times 9 \times 8 \times 7 = 5\,040$$How many if no digit may be repeated?
2. Factorials ($n!$)#
What is $n!$?#
$$n! = n \times (n-1) \times (n-2) \times \dots \times 2 \times 1$$| $n$ | $n!$ | Value |
|---|---|---|
| $0$ | Defined as $1$ | $1$ |
| $1$ | $1$ | $1$ |
| $2$ | $2 \times 1$ | $2$ |
| $3$ | $3 \times 2 \times 1$ | $6$ |
| $4$ | $4 \times 3 \times 2 \times 1$ | $24$ |
| $5$ | $5!$ | $120$ |
| $6$ | $6!$ | $720$ |
| $7$ | $7!$ | $5\,040$ |
| $10$ | $10!$ | $3\,628\,800$ |
Why $0! = 1$?#
There is exactly one way to arrange zero objects: do nothing. It also keeps the formula $n! = n \times (n-1)!$ consistent: $1! = 1 \times 0!$, so $0!$ must be $1$.
When Do We Use $n!$?#
When arranging all $n$ distinct items in a row, the number of arrangements is $n!$.
Why? First position: $n$ choices. Second: $n-1$. Third: $n-2$. … Last: $1$ choice. Total: $n!$.
Worked Example 4 — Arranging People#
$$6! = 720$$In how many ways can 6 people sit in a row?
3. Arrangements with Repeated Items#
If some items are identical, we divide by the factorial of each repeated group to avoid counting duplicate arrangements.
The Formula#
$$\text{Arrangements} = \frac{n!}{p! \times q! \times r! \times \dots}$$where $p, q, r, \dots$ are the frequencies of the repeated items.
Worked Example 5 — Letters with Repeats#
How many distinct arrangements of the letters in MISSISSIPPI?
Total letters: $11$
Repeated letters: S appears $4$ times, I appears $4$ times, P appears $2$ times
$$\frac{11!}{4! \times 4! \times 2!} = \frac{39\,916\,800}{24 \times 24 \times 2} = \frac{39\,916\,800}{1\,152} = 34\,650$$Worked Example 6 — Simpler Repeated Letters#
How many distinct arrangements of the letters in APPLE?
Total: $5$ letters. P appears $2$ times.
$$\frac{5!}{2!} = \frac{120}{2} = 60$$4. Constraint Problems#
Most exam questions add conditions that restrict the arrangement. Here are the key strategies.
Strategy 1: The “Block” Method (Items Must Be Together)#
If certain items must be adjacent, treat them as a single unit (“block”), arrange the blocks, then arrange the items within the block.
Worked Example 7 — Two People Must Sit Together#
5 people sit in a row. In how many ways can they sit if persons A and B must be next to each other?
Step 1: Treat A and B as one block → 4 “items” to arrange: $4! = 24$
Step 2: A and B can swap within their block: $2! = 2$
Total: $4! \times 2! = 24 \times 2 = 48$
Worked Example 8 — Three Letters Must Be Together#
How many arrangements of COUNTING have the letters N, T, I together?
Total letters in COUNTING: $8$ (C, O, U, N, T, I, N, G) — note N appears twice.
Treat {N, T, I} as one block → $6$ items to arrange (block + C, O, U, N, G).
Wait — one of the N’s is inside the block and one is outside. So:
Block arrangements (N, T, I internally): $3! = 6$
Remaining items to arrange: 6 items (block, C, O, U, N, G) but N still appears once outside:
$$\frac{6!}{1} \times 3! = 720 \times 6 = 4\,320$$Strategy 2: The “Fix First” Method (Items in Specific Positions)#
If items must be in specific positions (e.g., ends), fill those restricted positions first, then arrange the rest.
Worked Example 9 — Vowels at the Ends#
How many arrangements of $\{A, B, C, D, E\}$ have a vowel at each end?
Vowels: A, E (2 vowels)
Step 1 — Fill the ends: $2$ choices for the first end, $1$ remaining for the other end: $2 \times 1 = 2$
Step 2 — Fill the middle: 3 remaining letters in 3 positions: $3! = 6$
Total: $2 \times 6 = 12$
Strategy 3: The “Subtract” Method (Items Must NOT Be Together)#
If items must not be adjacent, calculate:
$$\text{Not together} = \text{Total arrangements} - \text{Together arrangements}$$Worked Example 10 — Two People NOT Together#
5 people sit in a row. A and B must NOT sit next to each other.
Total: $5! = 120$
Together (from Example 7): $4! \times 2! = 48$
Not together: $120 - 48 = 72$
5. Circular Arrangements#
When arranging $n$ items in a circle, there is no “first” position (rotating the whole arrangement doesn’t create a new one).
$$\text{Circular arrangements} = (n - 1)!$$Worked Example 11 — Round Table#
$$(6 - 1)! = 5! = 120$$In how many ways can 6 people sit around a circular table?
Why $(n-1)!$? Fix one person to remove the rotational symmetry. Then arrange the remaining $(n-1)$ people: $(n-1)!$.
6. Number Formation Problems#
These are common in CAPS exams: forming numbers with specific properties from given digits.
Worked Example 12 — Even Numbers#
How many 3-digit even numbers can be formed from $\{1, 2, 3, 4, 5\}$ without repetition?
An even number must end in an even digit. Available even digits: $\{2, 4\}$.
Step 1 — Fix the last digit: $2$ choices (2 or 4)
Step 2 — Fill the remaining positions: $4$ choices for the first digit, $3$ for the second.
Total: $4 \times 3 \times 2 = 24$
Worked Example 13 — Numbers Greater Than 400#
How many 3-digit numbers greater than 400 can be formed from $\{1, 2, 3, 4, 5\}$ without repetition?
The first digit must be $4$ or $5$ (to make the number $\geq 400$).
Step 1 — First digit: $2$ choices
Step 2 — Remaining digits: $4 \times 3 = 12$
Total: $2 \times 12 = 24$
🚨 Common Mistakes#
| Mistake | Why it’s wrong | Fix |
|---|---|---|
| Adding instead of multiplying | The counting principle uses multiplication for sequential choices | “AND” (do this AND that) = multiply |
| Forgetting to reduce choices | Without repetition, each slot has fewer options than the last | Track how many items remain after each choice |
| Not dividing for repeated items | APPLE has two P’s — swapping them doesn’t create a new arrangement | Divide by $p!$ for each repeated item |
| Block method — forgetting internal arrangements | The block {A, B} can be arranged as AB or BA | Always multiply by $k!$ for $k$ items in the block |
| Circular vs linear confusion | A circle has no “start” — $n!$ overcounts by a factor of $n$ | Use $(n-1)!$ for circular arrangements |
| Starting with the wrong slot | If one slot is restricted (must be even, must be specific digit), fill it first | Always handle constrained positions before free ones |
💡 Pro Tips for Exams#
1. Always Draw the Slots#
Before calculating, physically draw boxes for each position. Write the number of choices above each box. This prevents errors from trying to do everything in your head.
2. Handle Restrictions First#
If a question says “the first digit cannot be 0” or “vowels must be at the ends,” fill those restricted positions before anything else. The unrestricted positions are filled last.
3. The “Subtract” Strategy#
If a constraint is hard to count directly (e.g., “no two vowels adjacent”), it’s often easier to count the total and subtract the cases where the constraint is violated.
4. Check with Small Cases#
If you’re unsure about a formula, test it with a tiny example you can list by hand. For instance, arrangements of {A, B, C} = 6. Does your formula give 6? If yes, scale up.