120 lines
3.3 KiB
Markdown
120 lines
3.3 KiB
Markdown
---
|
|
title: Permutation Formula
|
|
---
|
|
## Permutation Formula
|
|
|
|
In probability, a permutation is an ordered set of elements created from a larger set of possible elements. In order to know the number of all possible permutations of a specific size that can be made from any given set of elements, we use the permutation formula which is as follows:
|
|
|
|
|
|
|
|
where *n* is the number of elements in the set of possible elements and *r* is the number of elements in the permutation of elements from the set of possible elements
|
|
|
|
To get the total number of possible permutations of size *r* from a set of size *n*, you first take the factorial (!) of *n*. Next you'll take the difference of *r* from *n* and then take the factorial of that difference. Then you divide the first factorial by the difference factorial to get the total number of possible permutations.
|
|
|
|
For example, if you wanted to know the total numbers of 4-digit number codes possible for a cell phone, you would first find the values for *n* and *r*.
|
|
|
|
Since the size of the code is 4 digits and the number of different numerals is 10, *n* = 10 and *r* = 4
|
|
Now we plug in *r* and *n* to the permutation formula:
|
|
|
|
|
|
|
|
Next, we take the difference of *n* and *r*:
|
|
|
|
|
|
|
|
Then, we take the factorial of both the numerator and denominator:
|
|
|
|
|
|
|
|
Finally, we divide each result to get the total number of possible 4 digit codes for a cell phone:
|
|
|
|
|
|
|
|
### Another Explanation
|
|
|
|
If I took a list of 3 color {red, blue, green}. How many ways could I arrange this?
|
|
|
|
{red, blue, green}
|
|
{red, green, blue}
|
|
|
|
{blue, red, green}
|
|
{blue, green, red}
|
|
|
|
{green, blue, red}
|
|
{green, red, blue}
|
|
|
|
In a list of 3 colors, I had 6 arrangements.
|
|
6 = 3! (3 X 2 X 1)
|
|
|
|
In another example, I take 4 letters {m, a, l, c} and arrange them in all the possible ways.
|
|
|
|
{m,a,l,c}
|
|
{m,a,c,l}
|
|
{m,l,c,a}
|
|
{m,l,a,c}
|
|
{m,c,a,l}
|
|
{m,c,l,a}
|
|
|
|
{a,m,l,c}
|
|
{a,m,c,l}
|
|
{a,l,c,m}
|
|
{a,l,m,c}
|
|
{a,c,m,l}
|
|
{a,c,l,m}
|
|
|
|
{l,m,a,c}
|
|
{l,m,c,a}
|
|
{l,a,c,m}
|
|
{l,a,m,c}
|
|
{l,c,m,a}
|
|
{l,c,a,m}
|
|
|
|
{c,m,a,l}
|
|
{c,m,l,a}
|
|
{c,a,l,m}
|
|
{c,a,m,l}
|
|
{c,l,m,a}
|
|
{c,l,a,m}
|
|
|
|
In total, that is 24 ways. 24 = 4! (4X3X2X1)
|
|
|
|
See a pattern?
|
|
|
|
In general, when asked how many ways can you arrange a list where order matters (meaning {1,2} != {2,1}), the formula is as follows:
|
|
|
|
n!, where n is the number of elements in the list.
|
|
|
|
Now, lets says we are asked how many ways arrange 2 out of the 4 letters.
|
|
|
|
{m,a,l,c}
|
|
|
|
{m,a}
|
|
{a,m}
|
|
|
|
{m,l}
|
|
{l,m}
|
|
|
|
{m,c}
|
|
{c,m}
|
|
|
|
{a,l}
|
|
{l,a}
|
|
|
|
{a,c}
|
|
{c,a}
|
|
|
|
{l,c}
|
|
{c,l}
|
|
|
|
That is 12 different ways.
|
|
|
|
When asked how many ways to arrange k elements from a list of n elements the formula is as follows:
|
|
|
|
n!/(n-k)!
|
|
|
|
So, from the example above, 4!/(4-2)! = 24/2 = 12.
|
|
|
|
#### More Information:
|
|
- [Helpful Khan Adcamedy video](https://www.khanacademy.org/math/precalculus/prob-comb/combinatorics-precalc/v/permutation-formula)
|
|
|