Permutation
Permutation - How it Works - Video
Why it Works
Why it Works:
Let's say we have a 4 digit PIN. How many options do we have for that PIN? Well, we could use the same number over and over, but that it isn't too smart. That would be with replacement. Since they are happening simultaneously, we multiply to get 10 * 10 * 10 * 10 = 10,000 outcomes.
Or we could use different numbers each time, which would be without replacement which is how most things work. Since they are happening simultaneously, we multiply to get 10 * 9 * 7 * 6 = 5,040 outcomes.
Now we get rewrite this using factorials. Remember, 10! is 10 * 9 * 7 *6 * 5 * 4 * 3 * 2 * 1. Our outcomes looks really similar to 10! except we don't have 6! So if divide 10! by 6!, we get our outcomes for the PIN without replacement. Now way we can write 6! is (10 - 4)! The 10 from the number of objects and the 4 from the number of positions.
So our formula is n! / (n - r)! where is n is the number of objects and r is the number of positions.
You might see that written as n P r or P (n, r), where is P is permutations.
Definition & Theorem
Definition & Theorem:
For permutations, we have a number of total elements or n be put in an arrangement with r elements without repetition.
To find the number of outcomes, we can use the theorem and multiply by the total elements or n and decrease by 1 until the desire amount of r elements.
And our formula is n! / (n - r)! where is n is the number of objects and r is the number of positions.
Example 1
Example 1:
Here we have P (6, 3). So we have 6 objects and 3 positions.
In keeping with the same PIN example we would have 6 options for the first number, 5 options for the second number, and 4 options for the third number. Since it happens simultaneously we multiply the numbers together. Our outcome is 120 different possibilities.
Or we could use the formula and we get 6! / ( 6 -3 )! ==> 6! / 3! = 6 * 5 * 4 = 120.
Example 2
Example 2:
Here we have P (2, 2). So we have 2 objects and 2 positions.
In keeping with the same PIN example we would have 2 options for the first number and 1 option for the second number. Since it happens simultaneously we multiply the numbers together. Our outcome is 2 different possibilities.
Or we could use the formula and we get 2! / ( 2 -2 )! ==> 2! / 0! = 2 * 1 = 2.