Permutation and Combination Calculator switches between P(n, r), for selections where order matters, and C(n, r), for selections where it does not, using the same n and r inputs. Both formulas share the same building block, n!, so the tool computes one factorial expansion and derives both results from it.
Tell permutations and combinations apart
The single question that separates the two is whether rearranging the chosen items produces a different outcome. Assigning first, second, and third place among 6 finalists cares about order, so it needs a permutation. Choosing 3 finalists to advance to a playoff round, with no ranking among them, does not care about order, so it needs a combination.
Compute the permutation count
Permutations use P(n, r) = n! / (n − r)!. For n = 6, r = 3, that is 6! / 3!, which is (6 × 5 × 4 × 3 × 2 × 1) / (3 × 2 × 1), simplifying to 6 × 5 × 4, equal to 120.
Permutation and Combination Calculator prints this cancellation directly so the factorial ratio does not need to be computed in full.
Compute the combination count
Combinations use C(n, r) = n! / (r! × (n − r)!). For the same n = 6, r = 3, that is 6! / (3! × 3!), which is 720 / (6 × 6), equal to 20.
Combinations divide the permutation count by r! because every group of r items can be arranged in r! different orders, and combinations treat all of those orderings as the same outcome.
See the relationship between the two formulas
P(n, r) always equals C(n, r) multiplied by r!, since a combination groups items and a permutation also orders them. For n = 6, r = 3: C(6, 3) × 3! = 20 × 6 = 120, matching P(6, 3) exactly.
This relationship lets Permutation and Combination Calculator derive either value once the other is known, without running two separate factorial computations.
Avoid this common mistake
Selecting the wrong formula for a word problem is the most frequent error in this topic. A lottery draw of 6 numbers from 49, where the numbers are reported as a set with no ranking, is a combination, C(49, 6).
Assigning 3 distinct prizes (first, second, third) to 3 of 10 entrants is a permutation, P(10, 3), because which entrant gets which specific prize matters. Read the problem for language like "in order," "ranked," or "assigned to a role" as signals for a permutation, and language like "a group of," "a subset," or "selected together" as signals for a combination.
Compute both values side by side
Entering n = 8, r = 2 into Permutation and Combination Calculator returns both results from the shared factorial base 8!. The permutation P(8, 2) = 8 × 7 = 56 counts ordered pairs, such as assigning a captain and a vice-captain from 8 players.
The combination C(8, 2) = 56 / 2! = 28 counts unordered pairs, such as choosing 2 players to represent the team with no distinction between them. The permutation is exactly 2! = 2 times the combination here, matching the general relationship for any n and r.
Build intuition with a shrinking checklist
A reliable way to decide between the two formulas is to picture assigning labels one at a time. If each selected item gets a distinguishable label, tag, or rank as it is chosen, the count needs a permutation. If the selected items are simply gathered into an unlabeled group with no distinction between them, the count needs a combination.
Working through a few small examples by hand, like choosing 2 letters from {A, B, C} both as an ordered pair and as an unordered pair, builds the intuition faster than memorizing the formulas alone.
Frequently asked questions
What is the difference between a permutation and a combination?
A permutation counts arrangements where order matters, while a combination counts selections where order does not matter. Permutations are always r! times larger than combinations for the same n and r.
How do you know whether to use a permutation or a combination?
To know whether to use a permutation or a combination, ask whether swapping the order of the chosen items creates a different outcome. If yes, use a permutation; if the items are treated as an unordered group, use a combination.
What is the formula that connects permutations and combinations?
The formula that connects them is P(n, r) = C(n, r) × r!, since a permutation also accounts for every ordering of the r selected items that a combination treats as identical.
How many ways can 3 winners be chosen from 10 entrants if order does not matter?
If order does not matter, 3 winners chosen from 10 entrants gives C(10, 3), which equals 120, since 10! / (3! × 7!) simplifies to (10 × 9 × 8) / 6.
How many ways can 3 distinct prizes be awarded to 10 entrants?
Awarding 3 distinct prizes to 10 entrants uses P(10, 3), which equals 720, since assigning first, second, and third place is an ordered selection: 10 × 9 × 8.
Can permutations and combinations use the same n and r?
Permutations and combinations can use the same n and r values, and Permutation and Combination Calculator computes both from a single pair of inputs, letting a comparison of the two results happen on one screen.
What is P(8, 2) and C(8, 2)?
P(8, 2) equals 56, computed as 8 × 7, while C(8, 2) equals 28, found by dividing the permutation count by 2!. The permutation is exactly twice the combination, since each unordered pair can be arranged in 2 different orders.
Is there a quick way to decide between a permutation and a combination?
A quick way to decide is to picture assigning a distinguishable label to each chosen item as it is selected: if that labeling matters to the outcome, use a permutation; if the items are only gathered into an unlabeled group, use a combination.
Summary
Permutation and Combination Calculator computes P(n, r) = n! / (n − r)! for ordered selections and C(n, r) = n! / (r! × (n − r)!) for unordered ones, from the same n and r.
The two formulas relate through P(n, r) = C(n, r) × r!. Read the word problem carefully for whether order matters before choosing which count answers the question.