Combination
Combination - How it Works - Video
Why it Works
Why it Works:
Let's say we have a 3 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.
Let's talk about the PIN for your ATM or your phone. You have 10 numbers for the first option, 9 numbers for the second option, and 8 numbers for the second option. To find the total number of outcomes we multiply the numbers since it happens simultaneously to get 720. Doe it matter the order we multiply these numbers? Will we get 720 each time? We have 6 ways to write 10 * 9 * 8.
10 * 9 * 8
10 * 8 * 9
9 * 10 * 8
9 * 8 * 10
8 * 9 * 10
8 * 10 * 9
If we multiply those numbers, every single one will give us 720. So the order doesn't matter. It likes a football team ; it doesn't matter the order you say the starting members, you are going to say each one time when you announce the starters. The same applies to a committee or your classmates.
So combination starts out the same as permutations, but we don't want the excess or the repeating scenarios. We need to divide by the number of positions or r to find the total without repeating scenarios.
So our formula for C (n, r) is P(n, r) / r! or n! / [ (n - r)! * r! ] where is n is the number of objects and r is the number of positions.
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 for C (n, r) is P(n, r) / r! or n! / [ (n - r)! * r! ] where is n is the number of objects and r is the number of positions..
Example 1
Example 1:
Here we have C (5, 2). So we have 5 objects and 2 positions.
In keeping with the same PIN example we would have 5 options for the first number and 4 options for the second number. Since it happens simultaneously we multiply the numbers together, but remember order doesn't matter so we need to get rid of the excess or repeating scenarios. We must divide by 2! because that is the number of positions. Our outcome is 10 different possibilities.
Or we could use the formula and we get 5! / [( 5 - 2 )! * 2!] ==> 5! / [3! * 2!] = 10.
Example 2
Example 2:
Here we have C (3, 3). So we have 3 objects and 3 positions.
In keeping with the same PIN example we would have 3 options for the first number, 2 options for the second number, and 1 option for the third number. Since it happens simultaneously we multiply the numbers together, but remember order doesn't matter so we need to get rid of the excess or repeating scenarios. We must divide by 3! because that is the number of positions. Our outcome is 1 different possibility.
Or we could use the formula and we get 3! / [( 3 - 3 )! * 3!] ==> 3! / [0! * 3!] = 1.
Example of Use
Example of Use:
We can use combinations to create this triangle. What triangle do think this is?