What combinations and permutations are
Combinations and permutations answer almost the same question with one decisive twist: from a set of n items, how many ways can you choose r? The difference is whether order counts. In a combination, order is irrelevant — picking Ann and then Ben gives the same group as picking Ben and then Ann. In a permutation, order matters — first and second place in a race are not interchangeable.
The factorial of a number, written n!, is the product of every integer from 1 up to n (for instance 5! = 5 × 4 × 3 × 2 × 1 = 120). It is the building block of both formulas, and on its own it counts how many ways you can arrange all n items in a row.
This calculator runs on BigInt, so it prints the exact whole number even when n is large and the answer has dozens of digits — no rounding and no overflow like the ones ordinary calculators hit.
How to use the calculator
- Enter the total number of available items in n.
- Enter how many you will choose in r. It must hold that
ris not greater thann. - Read the three cards: combinations, permutations, and
n!. They update as you type. - Press Copy results to grab all three numbers at once.
If r is greater than n, or you type something that is not a whole number, the calculator warns you instead of returning a wrong answer.
The formula
Both formulas start from the factorial:
- Combinations:
C(n, r) = n! / (r! · (n − r)!) - Permutations:
P(n, r) = n! / (n − r)!
The link between them is simple: P(n, r) = C(n, r) × r!. That makes sense, because each combination of r items can be reordered in r! different ways, and those reorderings are exactly what permutations count separately.
| Aspect | Combination | Permutation |
|---|---|---|
| Does order matter? | No | Yes |
| Formula | n! / (r!·(n−r)!) | n! / (n−r)! |
| Relationship | Always smaller or equal | P = C × r! |
| Typical example | Lottery numbers, a committee | Podium, password, ranking |
| C(5,2) vs P(5,2) | 10 | 20 |
Worked example
Picture a 6/49 lottery: six numbers are drawn from 49 and the order they come out in does not matter. That is a combination:
C(49, 6) = 49! / (6! · 43!) = 13,983,816
There are almost 14 million possible tickets, so the chance of matching all six with a single ticket is 1 in 13,983,816.
Now a case where order does count: in a race with 5 runners, how many ways can the gold, silver, and bronze podium be formed? We choose 3 and the order decides the medal, so it is a permutation:
P(5, 3) = 5! / 2! = 120 / 2 = 60
And if only gold and silver are handed out among 5 runners: P(5, 2) = 5! / 3! = 120 / 6 = 20, versus the C(5, 2) = 10 combinations. Exactly double, because r! = 2! = 2.
Frequently asked questions
What is the difference between a combination and a permutation?
In a combination the order of the chosen items changes nothing, so groups with the same members are counted only once. In a permutation each distinct order is a distinct outcome. That is why P(n, r) is always greater than or equal to C(n, r): the permutation counts all the reorderings that the combination lumps together.
Why is the result exact for large numbers?
The tool computes with BigInt, a number type that handles integers of arbitrary size. Calculators that rely on floating-point decimals lose precision past a certain magnitude; here the factorial and both formulas are solved with whole numbers, so the value shown is exact, digit for digit.
What happens if r is greater than n?
Choosing more items than exist is meaningless, so the calculator returns no number and shows a notice asking you to fix the inputs. In the formulas that case would require the factorial of a negative number, which is undefined.
What is the value of 0!?
By convention 0! = 1. It looks odd, but it keeps the formulas consistent: C(n, n) = n! / (n! · 0!) = 1, meaning there is exactly one way to choose all the items. Without 0! = 1 the arithmetic would not line up.
Can I use this for probability?
Yes — it is the first step. To get the probability of an event you divide favorable cases by possible cases, and combinations or permutations give you that total of possible cases. In a 6/49 lottery, for example, one ticket has a 1 in 13,983,816 chance.