Permutation and combination calculator

One question decides the formula: does the order matter? Everything else follows from that.

The question that decides it

Does order matter? That single test picks the formula.

Permutation — order matters. Gold, silver and bronze from ten runners. Picking A then B is a different outcome from B then A.
Combination — order does not matter. Three people for a committee from ten candidates. Choosing A then B gives the same committee as B then A.

A useful check on the language: a "combination lock" is misnamed. The order of the digits absolutely matters, so it is really a permutation lock.

The four formulas

nPr = n! / (n−r)! — ordered, no repeats. 10P3 = 720.
nCr = n! / (r!(n−r)!) — unordered, no repeats. 10C3 = 120.
n^r — ordered, repeats allowed. A 4-digit PIN is 10⁴ = 10,000.
(n+r−1)Cr — unordered, repeats allowed. Choosing 3 scoops from 10 flavours, repeats fine: 220.

The relationship between the first two is worth internalising: nPr = nCr × r!. Every unordered group of r items can be arranged r! ways, so permutations always exceed combinations by exactly that factor. 720 = 120 × 6.

Working an example

A lottery draws 6 numbers from 49, order irrelevant. That is 49C6 = 13,983,816. One ticket therefore has about a 1 in 14 million chance.

If the same lottery required the numbers in the drawn order, it would be 49P6 = 10,068,347,520 — 720 times harder, since 6! = 720. Same draw, radically different odds, decided entirely by whether order counts.

The symmetry of combinations

nCr always equals nC(n−r). Choosing 3 people from 10 to include is the same act as choosing 7 to exclude, so both give 120. This is not a coincidence but a genuine identity, and it is also a computational shortcut: to find 100C98, compute 100C2 = 4,950 instead.

This calculator uses that shortcut internally, along with incremental big-integer arithmetic rather than computing three enormous factorials and dividing. Naive implementations overflow at surprisingly small inputs; 200C100 is a 59-digit number but the factorials involved have hundreds of digits each.

Common questions

What is the difference between a permutation and a combination?

Permutations count ordered arrangements, combinations count unordered groups. Picking gold, silver and bronze is a permutation; picking three committee members is a combination. nPr is always at least nCr.

When do I use "with repetition"?

When the same item can be picked more than once. A PIN can repeat digits, so it is a permutation with repetition. Dealing cards from a deck cannot repeat, so it is not.

Why does nCr equal nC(n-r)?

Because choosing which r items to include is the same decision as choosing which n-r items to leave out. Selecting 3 from 10 to keep and selecting 7 to discard both give 120.

Can r be larger than n?

Only with repetition. Without it, you cannot choose 5 items from 3 distinct ones, so the calculator rejects it. With repetition allowed, picking 5 scoops from 3 flavours is perfectly valid.