MISSION · Data and probability
Counting Without Listing
Multiplying choices, and when order stops mattering
LEVEL 1 · EXPLORE
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LEVEL 2 · LEARN
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The idea in one sentence
If one choice has \(m\) options and the next has \(n\), the pair has \(m \times n\) outcomes — and almost all of counting is that rule, applied carefully.
Multiply the choices
Three shirts and four pairs of trousers make \(3 \times 4 = 12\) outfits. Not seven. Every shirt pairs with every pair of trousers, which is a grid, not a list.
This is the fundamental counting principle, and it extends as far as you need. A PIN of four digits, each 0 to 9, has \(10^4 = 10{,}000\) possibilities. A number plate of two letters and three digits has \(26^2 \times 10^3 = 676{,}000\).
The one thing to watch is whether the choices are independent. Picking a committee of two from five people isn’t \(5 \times 5\), because the second choice has only four people left. Once something is used up, the next factor shrinks.
Arrangements: order matters
How many ways can five books be arranged on a shelf? Five choices for the first slot, four for the next, then three, two, one:
\[5 \times 4 \times 3 \times 2 \times 1 = 120 = 5!\]That’s a factorial, and it counts the arrangements of a whole set. Factorials grow viciously — \(10!\) is over three million, \(20!\) exceeds the number of seconds since the pyramids were built.
If you’re arranging only some of them — 3 books from 8, in order — you stop the countdown early: \(8 \times 7 \times 6 = 336\). That’s a permutation.
\[P(n, r) = \frac{n!}{(n - r)!}\]Selections: order doesn’t matter
Choosing 3 pizza toppings from 8 is a different question. Ham, mushroom, olive is the same pizza as olive, ham, mushroom — so the 336 permutations have counted each pizza several times over.
How many times? Exactly \(3! = 6\), once for each ordering of the same three toppings. So divide it out:
\[C(n, r) = \frac{n!}{r!\,(n - r)!} \qquad \frac{336}{6} = 56\]That’s a combination. The whole difference between the two formulas is that one extra division by \(r!\), removing the orderings you didn’t want counted.
Which one to use
Ask a single question: would swapping two of the chosen items give a different outcome?
- Yes — a race podium, a PIN, seats in a row. Order matters, use permutations.
- No — a committee, a hand of cards, a set of toppings. Order doesn’t, use combinations.
Lottery draws are combinations, which is why the odds are so long. Choosing 6 numbers from 49 gives nearly 14 million equally likely selections, and no ordering of your ticket improves that.
Common mix-ups
- Adding when you should multiply. Separate stages multiply; genuinely exclusive alternatives add.
- Using permutations for a selection, which over-counts by \(r!\).
- Forgetting that choices shrink when items can’t repeat.
- Assuming \(0! = 0\). It’s 1 — there is exactly one way to arrange nothing.
Check yourself
A club of 10 people picks a president and a treasurer, then separately picks a committee of 2. How many ways for each?
President and treasurer are distinct roles, so order matters: \(10 \times 9 = 90\). The committee is a selection: \(\frac{90}{2} = 45\).
LEVEL 3 · REFLECT
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