We are going to take a break from proofs today to look at some methods of counting in math. Today we will cover Permutations.
A permutation is simply how many possible arrangements there are of any elements in a set. For example, let say we have a set with A,B,C,D in it. How many ways can we order ABCD?
ABCD, ADCB, ACDB, BADB,...and so on. Well, instead of listing every combination out we can use a permutation to count them. A permutation is defined by n factorial (n!)
n! = n(n-1)!
so for example 5!= 5 * 4 * 3 * 2 * 1
And for the ABCD problem we need to take 4! which is 4*3*2*1 = 24
So there are 24 possible permutations.
An interesting facet with permutations is that the numbers tend to rise quite quickly:
5!=120
6!=720
7!=5040
8!=40,320
9!=362,880
10!= 3,362,800
So if we tried to order ABCDEFGHIJ we would get over 3 million permutations!
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