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Special rules are used to count the number of possible outcomes of an experiment. These rules are used when dealing with large sets and sample spaces.
When the outcomes occur in a sequence with two different sets of outcomes, the multiplication rule is used.
If A is the count for the first set of outcomes and B is for the second set of outcomes then the:
Total Number of Outcomes = A x B
This can be demonstrated with a tree diagram:
| The total number of outcomes is: | A x B |
| 3 x 4 | |
| 12 | |
| A = 3 B = 4 |
Permutations are used to count the number of ordered arrangements. The length of the arrangement can be any length from 0 to the total number of elements in the set.
The formula used to calculate permutations is:
| P(n,k) = | nPk = |
n! (n-k)! |
In this case, P is for the permutation function and not for probability. n is the number of elements in the set and k is the length of the subset chosen to represent the arrangement.
Example:
given a set of five different objects, when two objects are chosen out of the five.
|
5! (5 - 2)! |
= | 20 Permutations |
When the set of total objects contains more than one object of the same type then counting the permutations is different.
The formula for finding the permutations when some of the objects are the same is:
|
n! (n1!n2!...nk!) |
n is the total number in the set and n1 is the number of elements of the first type while n2 is the number of elements of the second type all the way to k number of types.
Example:
If a basket contains two apples, three oranges and four bananas, then the fruit can be picked out of the basket:
|
9! (2!3!4!) |
= | 1260 ways |
When the subset of objects is taken from a larger set and the order in which these objects are arranged doesn't matter, the number of arrangements is called a combination.
To find the number of combinations by taking a number of objects from a set, the following formula is used:
| C(n,k) = |
|
= |
n! k!(n-k)! |
Where n is the number of total objects and k is the number of elements taken out.
Example:
If three letters are chosen from A, B, C, D, E and F then the number of combinations would be:
|
6! 3!(6-3)! |
= |
6! 3!3! |
= 20 |
| C(n,k) or |
|