1. Chapter 0.4
Counting Techniques

2. Vocabulary
Outcome: the result of a single trial of a process involving chance
Probability experiment: a process involving chance
Sample space: the set of all possible outcomes
Tree diagram: a diagram used to systematically list the outcomes in the sample space

3. Fundamental Counting Principle
This rule can be extended to three or more events as well.

4. Example 1. Fundamental Counting Principle
A bicycle manufacturer makes five- and ten-speed bikes in seven different colors and four different frame sizes. How many different bicycles does the manufacturer make?
There are two gear choices, seven color choices, and four frame choices. Apply the Fundamental Counting Principle:
2∙7∙4=56
Thus, 56 different bicycles can be made.

5. Counting problems can also involve determining the number of arrangements of objects.
Permutation: an arrangement of a group of distinct objects in a certain order
Example. There are six permutations of the letters A, B, and C:
ABC ACB BAC BCA CAB CBA
To determine this number mathematically, apply the Fundamental Counting Principle. There are three choices for the first letter, 2 choices for the second letter, and one option for the last letter.
3∙2∙1=6
The product 3∙2∙1 can also be written as 3! (three factorial).

6. The number of permutations of n distinct objects is n!
Factorial
The number of permutations of n distinct objects is n!

7. Example 2
There are 8 finalists in a band competition. In how many different ways can the bands be ranked if they cannot receive the same ranking?
8!=8∙7∙6∙5∙4∙3∙2∙1=40,320
So the band can be ranked 40,320 different ways

8. Permutations
Suppose you would rather know how many different ways the first, second, and third place rankings could be awarded.
This is the number of permutations of 8 finalists taken 3 at a time.
8∙7∙6=336
The number of permutations can also be found using factorials by finding the number of permutations of all 8 bands and dividing out the number of arrangements of bands that finish below third place

10. Example 3. Permutation of n Objects Taken r at a Time
How many different ways can two students be assigned to five tutors if only one student is assigned to each tutor?
Since the order in which the students are arranged determines the tutor to which a student is assigned, this situation calls for permutations.
5 𝑃 2 = 5! 5−2 ! = 5! 3!
= 5∙4∙3∙2∙1 3∙2∙1 =5∙4=20

11. Combinations
Combination: a selection of distinct objects in which the order of the objects selected does not matter.
Example. If we were to choose 3 out of 4 books (A, B, C, & D) for a report, choosing books A, B, & C is the same as choosing C, A, & B, or any other arrangement of these books.

12. Combinations of n Objects Taken r at a Time

13. Example 4. Combinations of n Objects Taken r at a Time
How many ways are there to choose 5 cards from a standard deck of 52 playing cards?
Since the order in which the cards are chosen is not important, this scenario uses combinations, rather than permutations
52 𝐶 5 = 52! 52−5 !5! = 52! 47!5! =2,598,960
So there are 2,598,960 ways to choose 5 cards from a standard deck of playing cards

15. Permutations or Combinations?
The first step in solving a counting problem is to determine whether order is important.

16. Example 5. Permutations or Combinations?
Twenty-five students write their names on slips of paper. Then three different names are chosen at random to receive prizes. Determine whether each situation involves permutations or combinations:
Choosing 3 people to receive a “no homework” coupon
Choosing 3 people to receive one of the following prizes: 1st prize, a new graphing calculator; 2nd prize, a “no homework” coupon; 3rd prize, a new pencil

17. Glencoe Algebra 2 Common Core textbook Page P12 #1 - 19
Assignment
Glencoe Algebra 2 Common Core textbook Page P12 #1 - 19