search
Search
Publish
menu
menu search toc more_vert
Robocat
Guest 0reps
Thanks for the thanks!
close
Comments
Log in or sign up
Cancel
Post
account_circle
Profile
exit_to_app
Sign out
help Ask a question
Share on Twitter
search
keyboard_voice
close
Searching Tips
Search for a recipe: "Creating a table in MySQL"
Search for an API documentation: "@append"
Search for code: "!dataframe"
Apply a tag filter: "#python"
Useful Shortcuts
/ to open search panel
Esc to close search panel
to navigate between search results
d to clear all current filters
Enter to expand content preview
icon_star
Doc Search
icon_star
Code Search Beta
SORRY NOTHING FOUND!
mic
Start speaking...
Voice search is only supported in Safari and Chrome.
Navigate to
A
A
share
thumb_up_alt
bookmark
arrow_backShare
Twitter
Facebook

Permutations

Probability and Statistics
chevron_right
Probability Theory
schedule Jul 1, 2022
Last updated
local_offer Arthur
Tags

Permutations refers to the number of ways of ordering $r$ elements from a total of $n$ elements.

The formula for permutations is as follows:

$$_nP_r = \frac{n!}{(n-r)!}$$

where:

$n$: total number of elements

$r$: number of elements chosen

Examples

Ordering 6 balls

If we had 6 balls numbered from 1 to 6, how many different ways could we order the 6 balls?

Here we are interested in ordering 6 balls from a total of 6 balls.

For the first ball, we can choose any of the 6 balls.

For the second ball, given we have already chosen 1 ball, we only have 5 balls to choose from.

For the last ball, given there is only 1 ball remaining, we only have 1 choice.

Hence this gives us:

$$\begin{align*} _6P_6 &= \frac{6!}{(6-6)!} \\ &= 6!\\ &= 6 * 5 * 4 * 3 * 2 *1 \\ &= 720 \end{align*}$$

So we have 720 different ways to order the 6 balls.

Ordering 3 balls from a total of 6

If we had 6 balls numbered from 1 to 6, how many different ways could we order exactly three of them?

Here we are interested in ordering three balls from a total of 6 balls.

For the first ball, we can choose any of the 6 balls.

For the second ball, given we have already chosen 1 ball, we only have 5 balls to choose from.

For the third and final ball, given we have already chosen 2 balls, we only have 4 balls left to choose from.

Using the formula this gives us:

$$\begin{align*} _6P_3&=\frac{6!}{(6-3)!}\\ &=\frac{6!}{3!}\\ &=\frac{6*5*4*3*2*1}{3*2*1}\\ &= 6*5*4 \\ &= 120 \end{align*}$$
robocat
Published by Arthur Yanagisawa
Edited by 0 others
Did you find this page useful?
thumb_up
thumb_down
Ask a question or leave a feedback...
thumb_up
0
thumb_down
0
chat_bubble_outline
0
settings
Enjoy our search
Hit / to insta-search docs and recipes!