Example 1
- n = 5
120
How many permutations of 5 items? P(5) = 120.
Type n. A permutation is an arrangement of n distinct items in a line. P(n) = n!. P(5) = 120. P(4) = 24. P(0) = 1, because the empty list has one way: arrange nothing.
The calculator uses the integer part of n. A negative n returns 0. This is not a combination and not a variation with k smaller than n.
Enter data and click Calculate.
A permutation of n items is one way to line all of them up. Five different books on a shelf have 120 orders, because P(5) = 5! = 120. Four people in a queue have 24 line-ups. The symbols P(n) and n! are the same number.
The formula is the factorial: you multiply 1·2·…·n. 0! = 1, so P(0) = 1. An empty list has nobody to shuffle, and the convention leaves one empty arrangement. 1! = 1: one person, one queue.
The field is n. The calculator uses the integer part: 5.7 counts as 5 and returns 120. A negative n in this implementation returns 0, not 1. On the factorial card a minus drops to 0! = 1. Different branch, different result on a minus.
A combination C(n,k) picks a subset with no order. A variation V(n,k) lines up only k of n. A permutation takes everyone: that is V(n,n). So P(4) = V(4,4) = 24.
The header units switch does not convert anything here. 5 stays 120 whatever the label says. This is a count of queues, not a length.
P(6) = 720. P(3) = 6. P(2) = 2. P(7) = 5040. P(10) = 3628800. P(0) and P(1) both return 1, for two different reasons.
P(n) = n! = 1 · 2 · … · n, and P(0) = 1
A permutation P(n) = n!. P(5) = 120. P(4) = 24. P(0) = 1. Arrangements of all n items, not a subset.
120
How many permutations of 5 items? P(5) = 120.
24
How many queues of 4 people? P(4) = 24.
1
Is P(0) zero? No. The empty list has one arrangement.
120. Five distinct things have 120 orders. That is the same 5! as on the factorial card.
An arrangement of all n distinct items. Order makes a new result. Anna-Bartek is not Bartek-Anna.
P(4) = 24. P(0) = 1. The empty list has one way: arrange nothing.
A permutation lines everyone up, so P(n) = V(n,n). A variation can take only k seats from n.
Not in this calculator. A negative n returns 0. On the factorial card a minus drops to 0! = 1.
No. The calculator uses the integer part, so 5.7 counts as P(5) = 120.
So the n! formula still holds at zero. One empty arrangement, not “no arrangements”.
P(6) = 720. P(3) = 6. P(2) = 2. Each next n multiplies by n.
No. A combination ignores order. A permutation keeps order and takes the whole set.
Five different people can stand in 120 orders. Two of five with no order is C(5,2) = 10, on combinations.
The calculator computes the same formula as the definition below.
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