Permutations & Combinations
“Order matters → permutation; order ignored → combination (C = P ÷ r!).”
The formula
ₙPᵣ = n!/(n−r)!, ₙCᵣ = n!/(r!(n−r)!)How to read it: a permutation counts ways to pick r items in order; a combination counts ways to pick r items ignoring order
- n
- — the total number of items
- r
- — how many you pick
- ₙPᵣ
- — permutations — count where order matters
- ₙCᵣ
- — combinations — count where order does not matter
The hook
'Who gets gold, silver, bronze' and 'which 3 reach the final' are counted differently — whether order matters is exactly what splits permutations from combinations.
In plain words
Picking r from n items: if the order you picked them in matters, it's a permutation (P); if only 'who got picked' matters, it's a combination (C).
The intuition
Pick them in an ordered line and you have permutations. But combinations don't care about order, so you must cancel the duplicates that are the same group in a different order (r! of them) — which is why a combination is a permutation divided by r!. C = P ÷ r!.
How it's built
P is 'n times (n−1) times … , r factors counting down'. C divides that P by r! to erase the ordering duplicates. The factorial n! multiplies 1 through n — the number of ways to line things up.
Example
From 5 people pick 2: permutations ₅P₂ = 5·4 = 20 (1st vs 2nd distinguished). Combinations ₅C₂ = 20 ÷ 2! = 20 ÷ 2 = 10 (just choosing 2).
Common mistake
Don't reach for combinations just because you see the word 'choose'. If the roles differ (president vs vice-president), it's a permutation; if there's no role distinction (just 2 reps), it's a combination.
Where it's used
Lottery odds, the number of possible passwords, team formation, card-game hands — the starting point of probability and statistics: 'how many ways are there?'
Where it came from
Systematic counting grew sharply from the letters between Pascal and Fermat as they reasoned about the odds in games of chance.
Prerequisites
Quick check
What is ₅C₂? (choosing 2 from 5)
- 10✓
- 20
- 25
- 60
Practice
Find ₄P₂ (arrange 2 out of 4 in order).
Answer: 12
- Multiply 2 factors counting down from 4: 4·3
- = 12
Takeaway: A permutation multiplies r factors down from n.
Find ₆C₂.
Answer: 15
- ₆C₂ = (6·5)/(2·1)
- = 30/2 = 15
Takeaway: A combination divides the permutation by r!.
Find 3!.
Answer: 6
- 3! = 3·2·1
- = 6
Takeaway: A factorial multiplies 1 up through the number.
Explain why ₙCᵣ = ₙPᵣ ÷ r! using 'choose 2 from 5'.
Answer: undefined
- The permutation ₅P₂=20 counts (A,B) and (B,A) separately
- But for choosing 2, those are the same group
- Divide out the 2!=2 orderings: 20÷2=10
Takeaway: A combination is a permutation with the ordering duplicates removed.