Permutations without Repetition

2. Permutations without Repetition

In this case, you have to reduce the number of available choices each time.

For example, what order could 16 pool balls be in? After choosing, say, number "14" you can't choose it again.

So, your first choice would have 16 possibilites, and your next choice would then have 15 possibilities, then 14, 13, etc. And the total permutations would be:

16 × 15 × 14 × 13 × ... = 20,922,789,888,000

But maybe you don't want to choose them all, just 3 of them, so that would be only:

16 × 15 × 14 = 3,360

In other words, there are 3,360 different ways that 3 pool balls could be selected out of 16 balls.

But how do we write that mathematically? Answer: we use the "factorial function"

So, if you wanted to select all of the billiard balls the permutations would be:

16! = 20,922,789,888,000

But if you wanted to select just 3, then you have to stop the multiplying after 14. How do you do that? There is a neat trick ... you divide by 13! ...

16 × 15 × 14 × 13 × 12 ...		= 16 × 15 × 14 = 3,360
13 × 12 ...

Do you see? 16! / 13! = 16 × 15 × 14

The formula is written:

where n is the number of things to choose from, and you choose r of them (No repetition, order matters)

Examples:

Our "order of 3 out of 16 pool balls example" would be:

16!	=	16!	=	20,922,789,888,000	= 3,360
(16-3)!		13!		6,227,020,800

(which is just the same as: 16 × 15 × 14 = 3,360)

How many ways can first and second place be awarded to 10 people?

10!	=	10!	=	3,628,800	= 90
(10-2)!		8!		40,320

(which is just the same as: 10 × 9 = 90)