Permutations & Combinations: P and C Formulas Explained

C(10,r)

120

210

252

210

120

Key Takeaways

Permutation (order matters): P(n,r) = n! ÷ (n−r)! — P(5,3) = 5×4×3 = 60

Combination (order irrelevant): C(n,r) = n! ÷ (r!(n−r)!) — C(5,3) = 10

The bridge: P(n,r) = C(n,r) × r! — every combination's members can be arranged r! ways

Repetition allowed: n^r — a 4-digit PIN has 10⁴ = 10,000 possibilities

Core answer: Order matters → permutations: P(n,k) = n!/(n−k)!. Order doesn't matter → combinations: C(n,k) = n!/(k!(n−k)!). Picking a president, secretary and treasurer from 10 people is P(10,3) = 720; picking a 3-person committee is C(10,3) = 120 — exactly 3! = 6 times fewer, because each committee has 6 internal orderings.

Which formula when

Situation	Order matters?	Formula	Example (n=10, k=3)
Podium / roles / password	Yes	P(n,k) = n!/(n−k)!	720
Committee / hand / toppings	No	C(n,k) = P/k!	120
PIN with repeats allowed	Yes, repeats ok	n^k	1,000
Arranging all n	Yes	n!	3,628,800

Worked examples

Example 1 — The 6/49 lottery. C(49,6) = 49!/(6!·43!) = 13,983,816. Your "lucky numbers" have the same 1-in-14-million chance as 1-2-3-4-5-6.

Example 2 — Poker pairs. Number of 5-card hands: C(52,5) = 2,598,960. Number of ways to get exactly one pair: choose rank (13) × 2 suits of 4 (C(4,2)=6) × 3 more ranks (C(12,3)=220) × suits (4³=64) = 1,098,240 → probability ≈ 42%.

Example 3 — Seating photos. 5 people line up: 5! = 120 arrangements. If two must stand adjacent, glue them into one unit: 4! × 2 = 48.

Example 4 — License plates. 2 letters + 4 digits, repeats allowed: 26² × 10⁴ = 6,760,000 plates per pattern — why cities add letters as they grow.

The symmetry tricks

C(n,k) = C(n,n−k): choosing 47 non-winners of 49 is the same count as choosing 6 winners. Compute the smaller side.
C(n,0) = C(n,n) = 1; C(n,1) = n.
Pascal's rule: C(n,k) = C(n−1,k−1) + C(n−1,k) — builds the triangle.

Common mistakes and myths

Using permutations for committees — if titles don't exist, order is irrelevant; divide by k!.
Forgetting the "with replacement" case — a 4-digit PIN is 10⁴ = 10,000, not P(10,4) = 5,040, because digits repeat.
Believing past draws change odds — each lottery draw is independent; last week's numbers are exactly as likely this week.
Double-counting in "at least one" problems — complement is easier: P(at least one) = 1 − P(none).
Thinking more tickets change the math meaningfully — 100 tickets in 6/49 is still a 0.0007% chance; combinatorics dwarfs purchases.

Frequently Asked Questions

How do I know whether to use P or C?

Ask: does the order of selection matter? Ranking gold/silver/bronze from 8 athletes — order matters → P(8,3) = 336. Picking 3 committee members from 8 — same 3 people regardless of listing order → C(8,3) = 56. Passwords, podiums, sequences → permutation. Groups, hands, subsets → combination. When unsure, ask "is ABC different from BAC?"

How do I calculate P(n,r)?

P(n,r) = n! ÷ (n−r)!, which is just r descending factors of n. P(5,3) = 5×4×3 = 60. P(10,2) = 10×9 = 90. P(n,n) = n! (arranging everything). Logic: first slot has n choices, second n−1, ..., r slots total — multiply. No formula memorization needed once you see the slots.

How do I calculate C(n,r)?

C(n,r) = n! ÷ (r!(n−r)!). C(5,3) = 120 ÷ (6×2) = 10. Useful symmetry: C(n,r) = C(n,n−r) — choosing 3 to include = choosing 2 to exclude, so C(5,3) = C(5,2) = 10. Compute the smaller side: C(100,98) = C(100,2) = 4,950. Calculator key: nCr.

How do I count when repetition is allowed?

Each slot independently has n choices: n^r. 4-digit PIN: 10⁴ = 10,000. 6-character lowercase password: 26⁶ ≈ 309 million. 8-character mixed-case + digits: 62⁸ ≈ 218 trillion. This is why length beats complexity for password strength — each added character multiplies the space by the alphabet size.

What are the odds of winning the lottery?

Classic 6/49: C(49,6) = 13,983,816 — about 1 in 14 million for the jackpot. 5/50 + 2/12 EuroMillions: C(50,5) × C(12,2) = 2,118,760 × 66 ≈ 1 in 140 million. Combinations because draw order does not matter. For perspective: you are more likely to be struck by lightning this year (~1 in 1.2M) than win most national jackpots.

What mistakes happen with P and C?

(1) Using P when the problem wants groups — "ABC and BAC are the same committee"; (2) forgetting to divide by r! when converting; (3) assuming no repetition when it is allowed (PINs, dice rolls: 6³ = 216, not P(6,3)); (4) the "combination lock" irony — it is actually a permutation lock since order matters; (5) computing n! directly for large n — cancel factors first: C(50,3) = (50×49×48)/(3×2×1) = 19,600.