Permutation & Combination Calculator — nPr, nCr, n!, n^r

Real-world use cases

• Compute lottery odds before you buy a ticket

  You want the real chance of matching all six numbers in a 6-from-49 draw. Enter n = 49, r = 6 and read the combinations row: C(49, 6) = 13,983,816. That is the number of equally likely tickets, so your odds of the jackpot on one line are 1 in 13,983,816. Steps: (1) set n to the size of the pool, (2) set r to how many numbers are drawn, (3) read C(n, r) — order does not matter in a lottery, so it is combinations, not permutations. Use the percentage-calculator afterwards to turn 1 / 13,983,816 into a percentage for a blog post.

• Size a password or PIN search space

  A security review asks "how many 8-character passwords are possible if each character is one of 95 printable ASCII symbols?". This is permutations WITH repetition because each position is independent and characters repeat. Enter n = 95, r = 8 and read n^r = 95^8 ≈ 6.6 × 10^15. Steps: (1) n is the alphabet size, (2) r is the length, (3) read the n^r row. Compare a 6-digit numeric PIN (n = 10, r = 6 → 1,000,000) to show stakeholders why length and alphabet matter more than "complexity rules".

• Count tournament seedings and round-robin orderings

  For a bracket you need the number of ways to award 1st, 2nd and 3rd place among 8 finalists. Order matters and no one wins twice, so it is permutations: enter n = 8, r = 3 → P(8, 3) = 336. Steps: (1) n = number of competitors, (2) r = number of ranked positions, (3) read P(n, r). For the full schedule of who-plays-whom, switch r to 2 and read C(8, 2) = 28 unique pairings in a round-robin.

• Show students why 30% + 10% off is not 40% off — and check their combinatorics homework

  A maths teacher assigns "in how many ways can 5 books be arranged on a shelf, and how many 3-book subsets exist?". Enter n = 5, r = 3: P(5, 3) = 60 arrangements, C(5, 3) = 10 subsets, and n! = 120 for the full-shelf arrangement. Steps: (1) read n! for arranging everything, (2) read P(n, r) for ordered sub-selections, (3) read C(n, r) for unordered ones. The worked-steps line under each result is copy-ready for the answer key, so marking is a paste rather than a recompute.

• Count ice-cream scoop or topping combinations for a menu

  A shop offers 6 flavours and lets customers pick 3 scoops, repeats allowed (three scoops of the same flavour is fine) and order on the cone does not matter. That is combinations WITH repetition: enter n = 6, r = 3 and read C(n + r − 1, r) = C(8, 3) = 56 distinct cones. Steps: (1) n = number of choices, (2) r = how many you pick, (3) read the C(n + r − 1, r) row when repeats are allowed but order is irrelevant. Useful for menu copy ("56 ways to build your cone") and inventory planning.

Common pitfalls

• Using permutations when you mean combinations. If you write the lottery jackpot odds as P(49, 6) you get 10,068,347,520 — that treats 1-2-3-4-5-6 as different from 6-5-4-3-2-1, which a lottery does not. Lotteries are unordered, so the correct count is C(49, 6) = 13,983,816. Ask yourself "does reordering my picks create a genuinely different outcome?" — if no, use C.

• Forgetting that "with repetition" changes which formula applies. A 4-digit PIN is NOT P(10, 4) = 5,040; it is 10^4 = 10,000, because digits can repeat. P(n, r) and C(n, r) both assume each item is used at most once. If your real-world problem lets the same item be chosen again, you need the n^r or C(n + r − 1, r) rows instead.

• Trusting a calculator that uses floating-point for large factorials. Many tools show 25! as `1.55e25` or round the last digits — that is lossy. For exact counts in cryptography, combinatorial proofs, or a probability denominator, an approximate factorial silently corrupts the answer. This tool uses BigInt, so 25! comes out exact to the final digit.

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