1. Essential Statistics 2E
William Navidi and Barry Monk

2. Counting
Section 4.4

3. Objectives
Count the number of ways a sequence of operations can be performed
Count the number of permutations
Count the number of combinations

4. Objective 1
Count the number of ways a sequence of operations can be performed

5. The Fundamental Principle of Counting
If an operation can be performed in 𝑚 ways, and a second operation can be performed in 𝑛 ways, then the total number of ways to perform the sequence of two operations is 𝑚𝑛. In general, if a sequence of several operations is to be performed, the number of ways to perform the sequence is found by multiplying together the numbers of ways to perform each of the operations.
Example: A certain make of automobile is available in any of three colors: red, blue, or green, and comes with either a large or small engine. In how many ways can a buyer choose a car? There are 3 choices of color and 2 choices of engine. The total number of choices is 3·2 = 6.

6. Example: Fundamental Principle of Counting
License plates in a certain state contain three letters followed by three digits. How many different license plates can be made? Solution: There are six operations in all; choosing three letters and choosing three digits. There are 26 ways to choose each letter and 10 ways to choose each digit. The total number of license plates is therefore 26·26·26·10·10·10 = 17,576,000.

7. Objective 2
Count the number of permutations

8. Permutations
The word “permutation” is another word for “ordering.” When we count the number of permutations, we are counting the number of different ways that a group of items can be ordered. The number of permutations of 𝒏 objects is 𝒏! Recall that 𝑛!=𝑛 𝑛−1 ⋯3(2)(1)

9. Example: Permutations
Five runners run a race. One of them will finish first, another will finish second, and so on. In how many different orders can they finish? The number of different orders the runners can finish is 5! = 5·4·3·2·1 = 120. Ten runners run a race. The first-place finisher will win a gold medal, the second-place finisher will win a silver medal, and the third-place finisher will win a bronze medal. In how many different ways can the medals be awarded? We use the Fundamental Principle of Counting. There are 10 possible choices for the gold medal winner. Once the gold medal winner is determined, there are nine remaining choices for the silver medal. Finally, there are eight choices for the bronze medal. The total number of ways the medals can be awarded is 10·9·8 = 720.

10. Permutation of 𝑟 Objects Chosen from 𝑛
In the last example, three runners were chosen from a group of ten, then ordered as first, second, and third. This is referred to as a permutation of three items chosen from ten. In general, a permutation of 𝑟 items chosen from 𝑛 items is an ordering of the 𝑟 items. It is obtained by choosing 𝑟 items from a group of 𝑛 items, then choosing an order for the 𝑟 items. The number of permutations of 𝑟 items chosen from 𝑛 is denoted 𝑛𝑃𝑟. The number of permutations of 𝒓 objects chosen from 𝒏 is 𝒏𝑷𝒓=𝒏∙ 𝒏−𝟏 ⋯ 𝒏−𝒓+𝟏 = 𝒏! 𝒏−𝒓 ! .

11. Example: Permutation of 𝑟 Objects from 𝑛
Five lifeguards are available for duty one Saturday afternoon. There are three lifeguard stations. In how many ways can three lifeguards be chosen and ordered among the stations? Solution: We are choosing three items from a group of five and ordering them. The number of ways to do this is 5𝑃3= 5! 5−3)! = 5! 2! = 5⋅4⋅3⋅2⋅1 2⋅1 =5⋅4⋅3=60

12. Objective 3
Count the number of combinations

13. Combinations
In some cases, when choosing a set of objects from a larger set, we don’t care about the ordering of the chosen objects; we care only which objects are chosen. For example, we may not care which lifeguard occupies which station; we might care only which three lifeguards are chosen. Each distinct group of objects that can be selected, without regard to order, is called a combination. The number of combinations of 𝒓 objects chosen from 𝒏 is 𝒏𝑪𝒓= 𝒏! 𝒓! 𝒏−𝒓 ! .

14. Example: Combination
At a certain event, 30 people attend, and 5 will be chosen at random to receive prizes. The prizes are all the same, so the order in which the people are chosen does not matter. How many different groups of 5 people can be chosen? Solution: Since the order of the 5 chosen people does not matter, we need to compute the number of combinations of 5 chosen from C5= 30! 5! 30−5)! = 30⋅29⋅28⋅27⋅26 5⋅4⋅3⋅2⋅1 =142,506

15. Permutations and Combinations on the TI-84
Calculator commands for permutations and combinations are accessed by pressing MATH and scrolling to the PRB menu.
The format for entering these commands is to enter the value of 𝑛 into the calculator, select either nPr (permutations) or nCr (combinations) from the PRB menu, and then enter the value of 𝑟. To run the command, press Enter.

16. Example: Counting in a Probability
A box of lightbulbs contains eight good lightbulbs and two burned-out bulbs. Four bulbs will be selected at random to put into a new lamp. What is the probability that all four bulbs are good? Solution: The outcomes in the sample space consist of all the combinations of four bulbs that can be chosen from 𝐶4= 10! 4! 10−4 ! = 3,628,800 24∙720 =210. The number of outcomes that correspond to selecting four good bulbs is the number of combinations of four bulbs that can be chosen from eight. 8𝐶4= 8! 4!∙ 8−4 ! = 40,320 24∙24 =70. 𝑃(Four good bulbs are selected) = = 1 3 .

17. You Should Know . . . How to use the Fundamental Principle of Counting
How to count the number of permutations of 𝑛 objects
How to count the number of permutations of 𝑟 objects chosen from 𝑛
How to count the number of combinations of 𝑟 objects chosen from 𝑛
How to compute permutations and combinations in the TI-84 PLUS calculator
How to compute probabilities using permutations and combinations