Permutations nPr Table Calculator

Enter Calculation

Number of objects (n)

Results

Total permutations of all n objects (nPn = n!)

265252859812191058636308480000000

for n = 30

r	nPr (permutations)
0	1
1	30
2	870
3	24360
4	657720
5	17100720
6	427518000
7	10260432000
8	235989936000
9	5191778592000
10	109027350432000
11	2180547008640000
12	41430393164160000
13	745747076954880000
14	12677700308232960000
15	202843204931727360000
16	3042648073975910400000
17	42597073035662745600000
18	553761949463615692800000
19	6645143393563388313600000
20	73096577329197271449600000
21	730965773291972714496000000
22	6578691959627754430464000000
23	52629535677022035443712000000
24	368406749739154248105984000000
25	2210440498434925488635904000000
26	11052202492174627443179520000000
27	44208809968698509772718080000000
28	132626429906095529318154240000000
29	265252859812191058636308480000000
30	265252859812191058636308480000000

What is the Permutations nPr Table Calculator?

This tool takes a single number of distinct objects, \(n\), and produces a complete table of permutations \({}_{n}P_{r}\) for every value of \(r\) from 0 up to \(n\). A permutation counts the number of ways to arrange \(r\) objects chosen from \(n\) when order matters. Because the results grow factorially, the calculator uses exact big-integer arithmetic so even very large values such as \(30!\) are shown precisely. This is pure combinatorics and applies identically everywhere; it is not region-specific.

How to use it

Enter the number of objects \(n\) (a non-negative whole number, default 30) and submit. The calculator returns the total number of arrangements of all objects (\({}_{n}P_{n} = n!\)) as the headline figure, plus a row-by-row table listing \({}_{n}P_{r}\) for \(r = 0, 1, 2, ..., n\). Read each row as "the number of ordered arrangements when choosing \(r\) of the \(n\) objects."

The formula explained

The defining formula is $$ {}_{n}P_{r} = \frac{n!}{(n - r)!} $$ Equivalently, \({}_{n}P_{r}\) is the falling factorial \(n \times (n-1) \times ... \times (n - r + 1)\), a product of exactly \(r\) descending terms. Special cases: \({}_{n}P_{0} = 1\) (the empty arrangement), \({}_{n}P_{1} = n\), and \({}_{n}P_{n} = n!\). The calculator builds the table efficiently by starting with \(P = 1\) and multiplying by \((n - r + 1)\) at each step, avoiding the need to compute huge factorials separately.

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nPr counts ordered arrangements of r items chosen from n distinct objects.

Worked example (n = 5)

Starting from 1 and multiplying down: \({}_{5}P_{0} = 1\), \({}_{5}P_{1} = 5\), $$ {}_{5}P_{2} = 5 \times 4 = 20 $$ $$ {}_{5}P_{3} = 5 \times 4 \times 3 = 60 $$ \({}_{5}P_{4} = 120\), and \({}_{5}P_{5} = 120\). Note that \({}_{5}P_{4}\) and \({}_{5}P_{5}\) are equal because the final factor is 1.

For n=5, ordered choices decrease 5, 4, 3, ... giving each nPr value.

FAQ

Why does nP0 equal 1? There is exactly one way to arrange zero objects: the empty arrangement.

What if r is larger than n? You cannot choose more objects than are available, so \({}_{n}P_{r} = 0\); the table therefore stops at \(r = n\).

How is nPr different from combinations nCr? Permutations count ordered arrangements, while combinations count unordered selections. They are related by $$ {}_{n}P_{r} = {}_{n}C_{r} \times r! $$