How this instrument works
"Without repetition" is the rule that once an item has been placed into one of the r positions, it is removed from the pool and cannot be placed again. That single rule is what makes the count shrink at every step: the first position has n items available, the second has n − 1 because one is already used, the third has n − 2, and so on. Multiplying that shrinking chain of choices together — or equivalently computing n! / (n − r)! — gives the total count of ordered arrangements, nPr.
It's worth naming the rule explicitly because the alternative is real and common. If repetition were allowed instead — an item could be placed in more than one position, like a digit reappearing in a PIN code — the pool would never shrink, every position would independently have the full n choices, and the count would follow n^r rather than n! / (n − r)!. The two formulas diverge fast: for 8 items arranged 3 at a time, without repetition gives 8 × 7 × 6 = 336 arrangements, while with repetition (if items could repeat) it would give 8³ = 512 — noticeably more, because repeats add outcomes a no-repeats rule forbids.
Physical, indivisible things almost always call for the without-repetition version: a person cannot stand on two podium steps at once, a physical book cannot occupy two shelf slots simultaneously, a runner cannot finish both first and second. Codes and symbols, by contrast, often allow repetition — nothing stops a 4-digit PIN from being 1111. Checking whether the same item could legitimately be reused across positions is the fastest way to pick the right formula.
- Enter the size of your full pool into n (total items) — the count of distinct items available before any are used up.
- Enter how many positions are being filled into r (items arranged, each used at most once).
- Read nPr — permutations without repetition beneath the inputs; the instrument computes n! / (n − r)! for you.
- If your scenario actually allows the same item to repeat across positions, use the permutation-with-repetition instrument instead — the count follows n^r, not n! / (n − r)!.
- If r exceeds n, the instrument flags it: without repetition, you cannot fill more positions than there are distinct items to fill them with.
Worked example — gold, silver, bronze among 8 finalists
Eight finalists compete, and gold, silver and bronze are awarded to three of them, with nobody allowed to win two medals. Enter n = 8 and r = 3. Because repetition isn't allowed, the pool shrinks by one at each step: 8 finalists could take gold, then 7 remain who could take silver (the gold winner is out), then 6 remain who could take bronze. Multiplying 8 × 7 × 6 gives 336, matching n! / (n − r)! = 8! / 5!. nPr reads 336.
If instead the rules allowed the same finalist to be recorded as the winner of more than one medal category — an unusual rule, but a useful contrast — each of the three medal picks would independently have all 8 finalists available, giving 8 × 8 × 8 = 8³ = 512 arrangements. The gap between 336 and 512 is exactly the effect of the no-repetition rule: it removes every arrangement where one finalist appears in more than one medal slot.
Questions
What does "without repetition" mean here?
It means each item, once used to fill one position, is removed from the pool and cannot be placed into a second position. Awarding gold to one finalist takes that finalist out of contention for silver — the pool of remaining candidates shrinks by one at every step, which is exactly what produces the n! / (n − r)! formula.
How is this different from the permutation-with-repetition calculator?
This instrument assumes each item can be used at most once, so the count shrinks at every position: n, then n − 1, then n − 2, and so on, giving n! / (n − r)!. The permutation-with-repetition instrument assumes items can be reused across positions — like a PIN digit repeating — so every position independently has all n choices, giving n^r instead. For n = 8, r = 3, that's 336 versus 512.
How do I know which formula applies to my situation?
Ask whether the same item could legitimately appear in more than one selected position. A person cannot occupy two podium spots, a physical book cannot sit on two shelf slots — those are without-repetition scenarios. A digit in a PIN code or a letter in a password can repeat freely — those need the with-repetition formula, n^r, instead.
What happens when r equals n?
The formula becomes n! / 0!, and since 0! = 1, that reduces to n! — every possible full ordering of the entire pool with no items left over. All 6 of 6 items, arranged without repetition, gives 6! = 720 orderings.
Can r be larger than n in a without-repetition permutation?
No. Without repetition, each item fills at most one position, so you cannot fill more positions than there are distinct items available. The instrument flags this rather than returning a result — with repetition allowed, by contrast, r can exceed n freely, since items are reused.
Is this the same formula as the plain permutation calculator?
Yes — nPr = n! / (n − r)! is identical either way; without-repetition is simply the explicit, spelled-out name for the standard permutation rule, useful when contrasting it directly against the with-repetition case (n^r) in the same discussion.