Find the number of ways to arrange r items out of a set of n, where the order matters.
Counts ordered selections — choosing a 1st, 2nd and 3rd place winner from a group is a permutation problem.
nPr: n! ÷ (n−r)!
A permutation calculator finds nPr, the number of ways to pick r items from n distinct items when the order of the picked items matters. Choosing a captain, vice-captain and wicket-keeper from 11 players is a permutation problem, because the same three people in different roles count as different outcomes.
Permutations are a Class 11 chapter and a regular source of questions in JEE, CAT, bank PO and other aptitude exams. They also appear whenever you design codes, passwords without repeated characters, race rankings or seating arrangements. The calculator takes n and r, multiplies the r largest factors of n!, and returns the count, which is often much bigger than people expect.
1. Identify n, the total number of distinct items available, and r, the number of positions to fill.
2. Confirm that order matters and that items cannot be repeated. If order does not matter, use combinations instead.
3. Check that r is between 0 and n.
4. Use nPr = n! ÷ (n − r)!.
5. In practice, multiply r consecutive numbers counting down from n: nPr = n × (n − 1) × … × (n − r + 1).
6. Check with a small case you can list by hand, such as 3P2 = 6: AB, BA, AC, CA, BC, CB.
There are r slots to fill in order. The first slot can take any of n items, the second any of the remaining n − 1, and the r-th slot any of the n − r + 1 items still unused. Multiplying gives n × (n − 1) × … × (n − r + 1). If you multiply and divide by (n − r)!, the product becomes n! ÷ (n − r)!. The factorial form is compact for algebra, but the countdown product is how you should actually calculate.
nP0 = 1, because there is exactly one way to choose nothing. nP1 = n, since you just pick one item. nPn = n!, because filling all n slots is the same as arranging everything. If r is larger than n there are no valid arrangements at all, because you run out of distinct items. The calculator shows 0 in that case, which is the correct count, although textbooks sometimes describe nPr as simply undefined there.
Every combination of r items can be ordered in r! ways, so nPr = nCr × r!. This is why permutations are always at least as large as combinations. If repetition is allowed, such as a 4-digit PIN where digits can repeat, the count is n^r instead, because every slot still has all n choices. This calculator handles the no-repetition case only. Arrangements with identical items, such as letters in a word, need division by the factorials of the repeated counts.
A school quiz has 12 finalists, and the organisers want to know in how many ways the gold, silver and bronze medals can be awarded.
Permutations: order matters: nPr = n! ÷ (n − r)! = 12! ÷ 9!
Multiply 3 numbers going down from 12: 12 × 11 × 10 = 1,320
Combinations: order does not matter, so divide by r!: nCr = nPr ÷ r! = 1,320 ÷ 6 = 220
Answer: Permutations, 12P3 1,320; Combinations, 12C3 220; Permutations if repeats are allowed 1,728
Using a permutation when order does not matter, such as choosing a committee, which overcounts by a factor of r!.
Using nPr for codes where characters may repeat. A 4-digit PIN from 10 digits has 10^4 possibilities, not 10P4.
Swapping n and r, for example entering n = 3 and r = 12.
Forgetting to handle identical items. The arrangements of BANANA are 6! ÷ (3! × 2!), not 6!.
Counting circular seating as a straight line. Around a round table, n people can be seated in (n − 1)! distinct ways.
Counting ways to award ranked prizes or positions in a competition.
Working out how many ordered schedules or batting orders are possible.
Estimating the number of passwords or codes with no repeated characters.
Solving Class 11, JEE and aptitude-test problems on arrangements.
Designing experiments where the order of treatments matters.
How is a permutation different from a combination?
A permutation cares about order — 1st, 2nd, 3rd place. A combination does not — it is just picking a group of 3, no matter the order. Permutations always give a bigger or equal count.