nPr Calculator computes P(n, r), the count of ordered arrangements of r items chosen from n distinct items without repetition. The formula is n! divided by (n − r)!, and the tool prints the factorial expansion alongside the final integer.
Compute P(n, r) directly
Permutations count arrangements where order matters, unlike combinations. Choosing a first-place, second-place, and third-place finisher from 8 runners is a P(8, 3) problem, since assigning the same three runners to different placements counts as different outcomes. P(8, 3) equals 8! / 5!, which simplifies to 8 × 7 × 6, equal to 336.
Expand the factorial ratio
n! / (n − r)! simplifies to the product of r consecutive integers counting down from n, because every term from (n − r)! downward cancels out of the full n! expansion. nPr Calculator shows this cancellation directly: P(8, 3) = (8 × 7 × 6 × 5 × 4 × 3 × 2 × 1) / (5 × 4 × 3 × 2 × 1) = 8 × 7 × 6.
Multiplying the three surviving factors gives 336 without needing to compute either factorial in full.
Work through a smaller example by hand
For P(5, 2), list the choices directly: 5 options for the first position, then 4 remaining options for the second, giving 5 × 4 = 20.
This matches the formula: 5! / 3! = 120 / 6 = 20. nPr Calculator uses this same countdown-multiplication method internally for any n and r, so the printed steps mirror what a hand count would produce for small cases.
Compare permutations to combinations
Permutations and combinations both select r items from n, but permutations multiply the combination count by r! to account for every ordering of the chosen items. P(n, r) = C(n, r) × r!.
For n = 8, r = 3, C(8, 3) = 56 and 3! = 6, and 56 × 6 = 336, matching P(8, 3) exactly.
Whenever a problem cares about assignment order, rank, or sequence, permutations are the correct count; when only the group membership matters, combinations are correct instead.
Avoid this common mistake
Using the combination formula for a problem that actually asks about order is the most common error with nPr problems. Assigning three distinct medals (gold, silver, bronze) to three of eight runners is an ordered assignment, so it needs P(8, 3) = 336, not C(8, 3) = 56.
The wrong formula understates the true count by a factor of r!, since it fails to distinguish which runner got which medal.
Work through a scheduling example
Assigning 3 different time slots to 3 of 7 available job applicants for interviews is an nPr problem, since each applicant gets a distinct, ordered slot.
Compute P(7, 3) = 7! / 4! = 7 × 6 × 5 = 210. nPr Calculator shows this as three descending factors starting at 7, matching the direct counting logic: 7 choices for the first slot, 6 remaining for the second, 5 remaining for the third.
Recognize order-sensitive language in a word problem
Word problems that need nPr typically describe assigning distinct titles, ranks, or time slots: "first, second, and third place," "president and vice president," "morning slot and afternoon slot." When a problem instead describes choosing an unordered group, such as "a committee of 3" or "a team of 5," with no distinct roles attached, nCr is the correct formula instead.
Reading for whether the chosen items get distinguishable labels is the fastest way to tell the two apart.
Frequently asked questions
What is the formula for nPr?
The formula for nPr is P(n, r) = n! / (n − r)!, where n is the total number of distinct items and r is the number chosen and arranged in order.
How do you calculate P(8, 3)?
To calculate P(8, 3), divide 8! by 5!, which simplifies to 8 × 7 × 6, equal to 336. The remaining factors of 5! cancel completely out of the full 8! expansion.
What is the difference between nPr and nCr?
nPr counts ordered arrangements and equals n! / (n − r)!, while nCr counts unordered selections and equals n! / (r! × (n − r)!). nPr is always r! times larger than nCr for the same n and r, since it also counts every ordering of the chosen items.
Can r be greater than n in nPr?
R cannot be greater than n in nPr without replacement, since there are not enough distinct items to fill more than n ordered positions. nPr Calculator returns an out-of-range message when r exceeds n.
What is P(n, n)?
P(n, n) equals n!, the full factorial of n, since arranging all n items in order uses every item with no items left over. P(4, 4) = 4! = 24.
What is P(n, 0)?
P(n, 0) equals 1 for any n, because there is exactly one way to arrange zero items: do nothing. This matches the convention that 0! equals 1.
What is P(7, 3)?
P(7, 3) equals 7! / 4!, which simplifies to 7 × 6 × 5, equal to 210 ordered arrangements of 3 items chosen from 7.
Why is nPr always larger than nCr for the same n and r?
nPr is always larger than nCr for the same n and r (when r is greater than 1) because nPr separately counts every ordering of the chosen items, while nCr groups all of those orderings into a single unordered selection. The exact ratio between them is r!.
Can nPr be computed without writing out the full factorials?
nPr can be computed without writing out the full factorials by multiplying only the r largest consecutive integers counting down from n, since every smaller factor cancels between the numerator and denominator in the ratio n! / (n − r)!.
Summary
nPr Calculator computes P(n, r) = n! / (n − r)! for ordered selections without repetition, showing the factorial cancellation as a product of r consecutive integers counting down from n. Enter n and r to get the exact permutation count, and compare against nCr when the problem does not care about the order of the selected items.