Permutations Calculator P(n,r)
Calculate P(n,r) = n!/(n-r)! — ordered arrangements of r items from n. Use when order matters: rankings, race finishes, passwords without repetition.
Formula & Reference
| Variable | Symbol | Formula | Units |
|---|---|---|---|
| Permutations Calculator P(n,r) | — | P(n,r) = n! / (n-r)! | positive integer |
Why Order Changes the Count
A permutation counts ordered selections. Choosing A then B is different from choosing B then A, so the number of arrangements of r distinct items selected from n distinct items is P(n,r)=n!/(n-r)!. The factorial ratio cancels the unused tail and leaves n(n-1)...(n-r+1), exactly r decreasing factors.
Permutations connect directly to combinations. A combination first selects an unordered group, and the r chosen objects can then be arranged in r! orders. Therefore P(n,r)=C(n,r)r!. This identity is a quick way to decide which count a problem needs: if changing the order creates a new outcome, use a permutation.
| Situation | Formula | Order? |
|---|---|---|
| Choose and arrange r from n | P(n,r)=n!/(n-r)! | Matters |
| Choose r from n | C(n,r)=n!/[r!(n-r)!] | Does not matter |
| r positions with repetition allowed | nr | Matters |
Special cases are useful checks: P(n,0)=1 because there is one empty arrangement, P(n,1)=n, and P(n,n)=n!. The calculator is intended for nonnegative integer n and r with r≤n.
Decision workflow: first ask whether the r selected positions are distinguishable. A gold-silver-bronze podium, president-vice president pair, or ordered password slot makes order matter. A committee or unordered subset usually does not. Then ask whether the same object can be reused. Standard P(n,r) assumes no repetition; if each of r positions independently allows any of n choices, the count is nr.
The factorial notation is compact, but multiplying only the r needed factors is often easier. P(12,3)=12×11×10=1,320; there is no need to compute 12! and 9! separately. This also makes the counting logic visible: 12 choices for the first position, then 11, then 10.
Permutation counts grow very quickly. P(20,10)=20×19×...×11, already a very large integer. In probability problems, the permutation count may appear in a denominator when all ordered outcomes are equally likely. Always define the sample space before turning a count into a probability.
Integer inputs matter. “5.7 objects” or “2.3 ranked places” has no ordinary combinatorial meaning, so n and r should be whole numbers. If a software tool rounds decimal entries automatically, the rounded problem is mathematically different from the one typed.
Step-by-Step Examples
P(8,3)
- 8×7×6=336 ways
P(10,4) no repeats
- 10×9×8×7=5,040
P(5,3)
- 5×4×3=60 finish orders
President and vice president: P(6,2)
- 6 choices for president × 5 remaining choices for vice president = 30.
P(10,0)
- The empty arrangement is one valid outcome, so the product of zero factors is 1.
Real-World Applications
Common Mistakes to Avoid
Use P when order matters.
All arrangements of n items.
n^r when repeats are allowed.
Permutation counts require whole numbers of objects and positions. Do not silently round 5.7 objects or 2.3 positions into a different counting problem.
Frequently Asked Questions
Related Math Calculators
Formula Explorer connections
Interpretation: This relationship counts discrete structures, analyzes logic and sets, or describes algorithmic growth. Assumption: Define the sample space, recurrence, logical variables and counting constraints precisely. Avoid double counting and distinguish worst, average and best cases.