Determining the Number of Possible Outcomes Formulas and Principles

MM1D1: Students will determine the number of outcomes related to a given event.

Fundamental Counting Principle Permutations and Combinations Math I Unit 4

Fundamental Counting Principle If one event can occur in m ways and another event can occur in n ways, then the number of ways that both events can occur is …

Example 1: You are buying a sandwich. You have a choice of 5 meats, 4 cheeses, 3 dressings, and 8 other toppings. How many different sandwiches with one meat, one cheese, one dressing, and one other topping can you choose?

Solution: You can use the fundamental counting principle to find the total number of sandwiches. Number of sandwiches = 5 x 4 x 3 x 8 = 480

Example 2: A town has telephone numbers that begin with 432 or 437 followed by four digits. How many different telephone numbers are possible?

Solution: The first three digits can only be 432 or 437…so there are only 2 choices. The last four digits have numbers 0 thru 9 to choose from….so there are 10 choices for each digit. 432 - ___ ___ ___ ___ 437 - ___ ___ ___ ___ 2 x 10 x 10 x 10 x 10 = 20,000 Phone number combinations

Extension: If you lived in the same town that had 432 and 437 as the first digits, but the last four digits could not repeat, how many different telephone numbers can be created? 2 choices for 432 or 437 10 choices for the first digit 9 choices for the second digit 8 choices for the third digit 7 choices for the fourth digit Solution: 2 x 10 x 9 x 8 x 7 = 10,080

Example 3: Twenty-six golfers are competing in the final round of a local competition. How many different ways can 3 of the golfers finish first, second, and third?

Solution: Any of the ____ golfers can finish first, then any of the ____ remaining golfers can finish second, and finally any of the ____ remaining golfers can finish third. So, the number of ways that the golfers can place first, second, and third is… 26 x 25 x 24 = 15,600

**Factorial: n! = n x (n - 1) x (n – 2) x (n – 3) x … x 1 Example: Find 5! 5 x 4 x 3 x 2 x 1 = 120 *** 0! Is always equal 1 ***

MM1D1b. Calculate and use simple permutations and combinations

PERMUTATIONS When finding the number of possible outcomes where the ORDER MATTERS, use a permutation. n is the total number r is the number taken at one time

Example 1: You were left a list of 8 chores. In how many orders can you complete 6 of the chores?

Solution: The number of permutations of 8 chores taken 6 at a time is:

Extension: How many different orders can you do all 8 chores?

Golfers Revisited… Twenty-six golfers are competing in the final round of a local competition. How many different ways can 3 of the golfers finish first, second, and third?

Solution: There are 26 golfers to choose from. Since three will be chosen … and the award that they receive MATTERS… it can be solved using permutations. 26 P 3 = 15, 600

Combinations When finding the number of possible outcomes where the order DOES NOT matter, use a combination. n is the total number r is the number taken at one time

Example 1: Eighteen basketball players are competing for 5 starting positions. The players selected to start will make up the first team. If the order in which the players are selected is not important, how many different first teams are possible?

Solution: The number of ways to choose 5 players from 18 is:

On the Calculator: TI–84 Plus: 18 MATH PRB (3) nCr 5 ENTER 8568 TI-30XIIS: 18 PRB nCr ENTER 5 8568

Example 2: Your English teacher has asked you to select 3 novels from list of 10 to read as an independent project. In how many ways can you choose which books to read?

Solution: Since you have 10 novels to choose from and you are choosing only 3…

MM1D1a. Apply the addition and multiplication principles of counting

Deciding Whether to Add or Multiply Addition Principle: “OR” If two or more events cannot occur at the same time…ADD the combinations together Multiplication Principle: “AND” If two or more events can occur at the same time…MULTIPLY the combinations together

Example 1 Part A: A restaurant serves omelets that can be ordered with any of the ingredients shown: Vegetarian: green pepper, red pepper, onion, mushroom, tomato, cheese Meat: ham, bacon, sausage, steak A. Suppose you want exactly 2 vegetarian ingredients and 1 meat ingredient in your omelet. How many different types of omelets can you order?

Solution: You can choose 2 of 6 vegetarian ingredients and 1 of 4 meat ingredients. So, the number of possible omelets is: Since you are choosing both for the same omelet, you will multiply the combinations together.

Example 1 Part B: B. Suppose you can afford at most 3 ingredients in your omelet. How many different types of omelets can you order?

Solution: You can order an omelet with 0, 1, 2, or 3 ingredients. Because there are 10 items to choose from… You will either be choosing 0 OR 1 OR 2 OR 3 - you will add the combinations.

Example 2: A movie rental business is having a special on new releases. The new releases consist of 8 comedies, 3 family, 10 action, 7 dramas, and 2 mystery movies. Part 1: Suppose you want exactly 3 comedies and 2 dramas. How many different movie combinations can you rent?

Solution: You can choose 3 of the 8 comedies and 2 of the 7 dramas. So, the number of possible movie combinations is: 56 x 21 = 1176

Part 2: Suppose you can afford at most 2 movies. How many movie combinations can you rent (assuming that you do rent a tleast 1 movie)?

Solution: Since you can either get 1 movie OR 2 movies…