Permutations Calculator
Calculate the number of permutations P(n, r) — arranging r items from n where order matters. See also Combinations Calculator and Factorial Calculator.
How to Calculate Permutations
A permutation is an arrangement of items where order matters. To calculate P(n, r), divide n! by (n−r)!. For example, arranging 3 people in a line from a group of 10 — the order they stand in matters. The first position has 10 choices, the second has 9, and the third has 8, giving 10 × 9 × 8 = 720 permutations.
Permutations Formula
P(n, r) = n! / (n − r)!
Alternatively:
P(n, r) = n × (n-1) × (n-2) × ... × (n-r+1)
Relationship to combinations:
P(n, r) = C(n, r) × r!
Example
P(10, 3) = 10! / 7!
= 10 × 9 × 8
= 720
Compare with C(10, 3) = 120
P(10, 3) = C(10, 3) × 3! = 120 × 6 = 720 ✓
Frequently Asked Questions
When should I use permutations vs combinations?
Use permutations when the order of selection matters (rankings, passwords, seating arrangements). Use combinations when order does not matter (committees, lottery numbers, card hands).
What is P(n, n)?
P(n, n) = n!, which is the number of ways to arrange all n items. For example, P(4, 4) = 4! = 24 ways to arrange 4 items in a line.
What is P(n, 1)?
P(n, 1) = n. There are n ways to choose 1 item from n items when order matters (which is the same as combinations in this case).
How are permutations used in real life?
Permutations are used in password security (counting possible passwords), sports (race finishing orders), scheduling (task ordering), and cryptography.
Solved Examples — Permutations
Example: How many different 4-digit PIN codes can be made from digits 0–9 (no repeats)?
Solution:
Step 1: Order matters, no repetition → permutation
Step 2: P(10, 4) = 10! / (10−4)! = 10! / 6!
Step 3: = 10 × 9 × 8 × 7 = 5,040
Answer: 5,040 possible PINs (without repetition)
Example: In a race of 12 runners, how many ways can gold, silver, and bronze be awarded?
Solution:
Step 1: Order matters (1st ≠ 2nd ≠ 3rd)
Step 2: P(12, 3) = 12 × 11 × 10 = 1,320
Answer: 1,320 different outcomes
Example: How many ways can the letters in "MATH" be arranged?
Solution:
Step 1: 4 distinct letters, arrange all → P(4, 4) = 4!
Step 2: 4! = 4 × 3 × 2 × 1 = 24
Answer: 24 arrangements
Example: How many ways can 5 different books be arranged on a shelf using only 3 spots?
Solution:
Step 1: P(5, 3) = 5! / (5−3)! = 5! / 2!
Step 2: = (5 × 4 × 3 × 2!) / 2! = 5 × 4 × 3 = 60
Answer: 60 arrangements
Practice Questions
Try these on your own:
- Calculate P(8, 3) (Answer: 336)
- How many 3-letter "words" can be formed from A-Z (no repeats)? (Answer: P(26,3) = 15,600)
- How many ways can 7 people sit in a row of 7 chairs? (Answer: 7! = 5,040)
- A license plate has 3 letters then 4 digits (repeats allowed). How many plates? (Answer: 26³ × 10⁴ = 175,760,000)
- From 9 candidates, how many ways can president and vice president be chosen? (Answer: P(9,2) = 72)
- How many distinct arrangements of the word "BOOK"? (Answer: 4!/2! = 12, due to repeated O)
Common Mistakes to Avoid
The most common mistake is using permutations when combinations should be used. Remember: permutations are for ARRANGEMENTS where order matters (ABC ≠ BAC), combinations are for SELECTIONS where order doesn't matter (ABC = BAC). Another error is forgetting about repetition — if items can repeat (like digits in a password), the formula changes to n^r instead of P(n,r). For words with repeated letters, divide by the factorial of each repeated letter count (e.g., MISSISSIPPI has 11!/(4!×4!×2!) arrangements). Students also sometimes confuse P(n, n) = n! with P(n, r) for r < n. Finally, remember that r must be ≤ n — you cannot arrange 5 people in 8 seats (it would be P(8, 5) with 8 seats and 5 people).
Key Takeaways
- Permutations: P(n, r) = n! / (n−r)!. ORDER MATTERS.
- P(n, n) = n! (arranging all n items). P(n, 1) = n (choosing 1 from n).
- With repetition allowed: n^r possibilities (e.g., 4-digit PIN with repeats = 10⁴ = 10,000).
- For repeated items: divide n! by the product of factorials of each repeated element.
- Relationship: P(n, r) = C(n, r) × r! — permutations are always ≥ combinations.
- Applications: passwords, seating arrangements, race results, scheduling, and license plates.