CK-12 Probability and Statistics - Advanced

3.6. Basic Counting Rules http://www.ck12.org

( 3 )( 5 )( 4 ) = 60

There are 60 different dinners that can be ordered by the customers.

Example:

Here is a classic example. In how many different ways can you seat 8 people at a dinner table?

Solution:

For the first seat, there are eight choices. For the second, there are seven remaining choices, since one person has
already been seated. For the third seat, there are 6 choices, since two people are already seated. By the time we get
to the last seat, there is only one seat left. Therefore, using the multiplicative rule above, we get

( 8 )( 7 )( 6 )( 5 )( 4 )( 3 )( 2 )( 1 ) = 40 , 320

The multiplication pattern above appears so often in statistics that it has its own name and its own symbol. So we
say “eight factorial,” and we write 8!.

Factorial Notation

n!=n(n− 1 )(n− 2 )... 1

Example:

Suppose there are 30 candidates that are competing for three executive positions. How many different ways can you
fill the three positions?

Solution:

This is a more difficult problem than the examples above and we will use the second version of the Multiplicative
Rule of Counting. We need to analyze it in the following way:

The executive positions can be denoted byk=3 sets of elements that correspond to

n 1 =The number of candidates that are available to fill the first position

n 2 =The number of candidates remaining to fill the second position

n 3 =The number of candidates remaining to fill the third position

Hence,

n 1 = 30

n 2 = 29

n 3 = 28

The number of different ways to fill the three positions is

n 1 n 2 n 3 = ( 30 )( 29 )( 28 ) = 24 ,360 possible positions.

The arrangement of elements indistinct order, as the example above shows, is called thepermutation.Thus, from
the example above there are 24,360 possiblepermutationsof three positions drawn from a set of 30 elements.