# Permutations Calculator nPr

## Calculator Use

Like the
Combinations Calculator the
**Permutations Calculator** finds the number of subsets that can be taken from a larger set. However, the order of the subset matters. The Permutations Calculator finds the number of subsets that can be created including subsets of the same items in different orders.

- Factorial
- There are n! ways of arranging n distinct objects into an ordered sequence, permutations where n = r.
- Combination
- The number of ways to choose a sample of r elements from a set of n distinct objects where order does not matter and replacements are not allowed.
- Permutation
- The number of ways to choose a sample of r elements from a set of n distinct objects where order does matter and replacements are not allowed. When n = r this reduces to n!, a simple factorial of n.
- Combination Replacement
- The number of ways to choose a sample of r elements from a set of n distinct objects where order does not matter and replacements are allowed.
- Permutation Replacement
- The number of ways to choose a sample of r elements from a set of n distinct objects where order does matter and replacements are allowed.
- n
- the set or population
- r
- subset of n or sample set

## Permutations Formula:

For n ≥ r ≥ 0.

Calculate the permutations for P(n,r) = n! / (n - r)!. "The number of ways of obtaining an ordered subset of r elements from a set of n elements."[1]

### Permutation Problem 1

**Choose 3 horses from group of 4 horses**

In a race of 15 horses you beleive that you know the best 4 horses and that 3 of them will finish in the top spots: win, place and show (1st, 2nd and 3rd). So out of that set of 4 horses you want to pick the subset of 3 winners and the order in which they finish. How many different permutations are there for the top 3 from the 4 best horses?

For this problem we are looking for an ordered subset of 3 horses (r) from the set of 4 best horses (n). We are ignoring the other 11 horses in this race of 15 because they do not apply to our problem. We must calculate P(4,3) in order to find the total number of possible outcomes for the top 3 winners.

P(4,3) = 4! / (4 - 3)! = **24 Possible Race Results**

If our 4 top horses have the numbers 1, 2, 3 and 4 our 24 potential permutations for the winning 3 are {1,2,3}, {1,3,2}, {1,2,4}, {1,4,2}, {1,3,4}, {1,4,3}, {2,1,3}, {2,3,1}, {2,1,4}, {2,4,1}, {2,3,4}, {2,4,3}, {3,1,2}, {3,2,1}, {3,1,4}, {3,4,1}, {3,2,4}, {3,4,2}, {4,1,2}, {4,2,1}, {4,1,3}, {4,3,1}, {4,2,3}, {4,3,2}

### Permutation Problem 2

**Choose 3 contestants from group of 12 contestants**

At a high school track meet the 400 meter race has 12 contestants. The top 3 will receive points for their team. How many different permutations are there for the top 3 from the 12 contestants?

For this problem we are looking for an ordered subset 3 contestants (r) from the 12 contestants (n). We must calculate P(12,3) in order to find the total number of possible outcomes for the top 3.

P(12,3) = 12! / (12-3)! = **1,320 Possible Outcomes**

### Permutation Problem 3

**Choose 5 players from a set of 10 players**

An NFL team has the 6th pick in the draft, meaning there are 5 other teams drafting before them. If the team believes that there are only 10 players that have a chance of being chosen in the top 5, how many different orders could the top 5 be chosen?

For this problem we are finding an ordered subset of 5 players (r) from the set of 10 players (n).

P(10,5)=10!/(10-5)!= **30,240 Possible Orders**

## References

[1] For more information on permutations and combinations please see Wolfram MathWorld: Permutation.