The Fundamental Counting Principle
You have 3 shirts and 4 pairs of trousers.
How many different outfits can you put together? You could list them all — but there's a
shortcut so reliable it has a grand name: the fundamental counting principle.
It says something wonderfully simple.
If one choice can be made in m ways and a second, independent choice
in n ways, then the two together can be made in
m \times n \text{ ways.}
Just multiply. Three shirts and four trousers give
3 \times 4 = 12 outfits. How many 4-digit
PINs are there? Ten choices for each of four digits, so
10 \times 10 \times 10 \times 10 = 10\,000. The principle chains on
for as many stages as you like — keep multiplying the number of ways at each step and you're done.
Why it works: branches of a tree
Why multiply, rather than add? A tree diagram shows the reason at a glance. Each new choice splits
every existing branch into more branches, so the number of leaves at the tips
multiplies. Step through it below: two shirts, each pairing with two bottoms,
grows to four complete outfits at the tips — 2 \times 2 = 4.
Add a third stage — say two pairs of shoes — and every one of those four branches splits in two
again: 4 \times 2 = 8 leaves. That relentless doubling (or tripling, or
tenfold-ing) at each stage is the whole idea.
A worked example: choosing a three-course meal
A restaurant offers 4 starters, 5 mains and
3 desserts. How many different three-course meals could you order?
Three independent stages, so multiply all three counts:
4 \times 5 \times 3 = 60 \text{ meals.}
Sixty — far more than you'd guess, and you'd never want to draw that tree by hand. That is the
power of the principle: it counts an enormous sample space without ever listing it.
Repetition, and arrangements
Each stage just needs its own count of options — and whether items can repeat
changes that count. A 4-digit PIN lets every digit be
0–9 again and again, so the count
stays at 10 the whole way:
10 \times 10 \times 10 \times 10 = 10^4 = 10\,000 \text{ PINs.}
When items cannot repeat, the count shrinks at each stage, because every choice
uses one up. Arranging 5 different books in a row uses
5 options for the first slot, then only 4
books left, then 3, and so on:
5 \times 4 \times 3 \times 2 \times 1 = 120.
- independent choices multiply:
m \times n \times \dots ways in total;
- with repetition allowed, each of k stages has
the same count s, giving s^k;
- without repetition, the count drops by one each stage
(n, n-1, n-2, \dots);
- arranging all n distinct items in order gives
n! = n \times (n-1) \times \dots \times 1.
Repeat or not? Two everyday cases side by side
The same principle handles both — the only question is whether a choice can be reused.
-
A padlock code (repetition allowed). Three dials, each
0–9, and the same digit may appear twice:
10 \times 10 \times 10 = 1000 codes.
-
A race podium (no repetition). With 8 runners, the
same person can't take two places, so the count falls each stage —
8 for gold, then 7 for silver, then
6 for bronze:
8 \times 7 \times 6 = 336 podiums.
Both are "three stages, multiply", yet the padlock keeps 10 options
throughout while the podium loses one runner each time. Spotting which case you're in is the
whole skill.
A real one: how many number plates?
Modern UK number plates look like AB12 CDE: two letters, two digits, then three
more letters. How many are possible? Work stage by stage and multiply, remembering that letters
can repeat (26 each) and digits can repeat (10
each):
26 \times 26 \times 10 \times 10 \times 26 \times 26 \times 26 \approx 1.2 \text{ billion.}
Over a billion plates from just seven stages — which is exactly why the scheme has lasted so long
without running out. You would never list them, but the counting principle sizes the whole
sample space in a single line of multiplication.
The one mistake everybody makes: they add when they should
multiply. Three shirts and four trousers give
3 \times 4 = 12 outfits — not
3 + 4 = 7. Adding leaves you with a wildly wrong (and far too small)
answer.
Here is the rule that keeps you straight — the AND / OR test:
-
AND → multiply. Combining one from each stage — "a shirt
and trousers" — multiplies: 3 \times 4.
-
OR → add. Picking just one thing from separate groups — "a shirt
or a hat" from 3 shirts and 2
hats — adds: 3 + 2 = 5 choices.
Ask yourself "am I combining one from each, or choosing a single thing?" — that
AND-multiply / OR-add distinction is the crux of nearly every counting question.
Each extra digit doesn't add to the possibilities — it multiplies them
by 10. A 4-digit PIN has
10\,000 combinations; a 6-digit one has
10^6 = 1\,000\,000 — a whole million. That explosive
growth is the entire basis of password security: every character you add makes the code
exponentially harder to crack.
This idea grows straight into permutations and combinations — the maths of how
many ways things can be arranged or chosen. That's what powers the odds of the National Lottery
(choosing 6 numbers from 59 gives about
45 million tickets, which is why the jackpot is so rare) and a huge
slice of probability besides. It all begins with one modest instruction: multiply.
See it explained