Section 4.3 Basic Counting Rules HAWKES LEARNING SYSTEMS math courseware specialists Copyright © 2008 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved.

For a sequence of n experiments where the first experiment has k 1 outcomes, the second experiment has k 2 outcomes, the third experiment has k 3 outcomes, and so forth. The total number of possible outcomes for the sequence of experiments is k 1 ∙ k 2 ∙ k 3 … k n. Probability, Randomness, and Uncertainty 4.3 Basic Counting Rules HAWKES LEARNING SYSTEMS math courseware specialists Fundamental Counting Principle:

Suppose you are ordering a banana split with 4 scoops of ice cream. If there are 25 flavors to choose from, and you want each scoop to be a different flavor, then how many different ways can this banana split be made? Calculate the total number of outcomes: HAWKES LEARNING SYSTEMS math courseware specialists Probability, Randomness, and Uncertainty 4.3 Basic Counting Rules There are 4 slots to fill, one for each different scoop of ice cream. For the first scoop you, will have 25 flavors to choose from. For the second scoop, you will have 24 flavors to choose from, and so forth. Therefore there are (25)(24)(23)(22) = 303,600 different ways to make your banana split. Solution:

The product of all positive integers less than or equal to n. Symbolically, n! = n(n – 1)(n – 2)…(2)(1) 0! = 1 Probability, Randomness, and Uncertainty 4.3 Basic Counting Rules HAWKES LEARNING SYSTEMS math courseware specialists Factorial:

a.7! 7!  (7)(6)(5)(4)(3)(2)(1) b. c. d. Calculate the following factorial expressions: HAWKES LEARNING SYSTEMS math courseware specialists Probability, Randomness, and Uncertainty 4.3 Basic Counting Rules  5040

When you need to count the number of ways objects can be chosen out of a group, in which the order the objects are chosen is important, it is called a permutation. When you need to count the number of ways objects can be chosen out of a group, in which the order the objects are chosen is not important, it is called a combination. Probability, Randomness, and Uncertainty 4.3 Basic Counting Rules HAWKES LEARNING SYSTEMS math courseware specialists Permutations and Combinations: “n things permuted r at a time” “n choose r ”

A class of 23 fifth graders is holding elections for class president, vice-president, and secretary. How many different ways can the officers be elected? Answer the following: HAWKES LEARNING SYSTEMS math courseware specialists Probability, Randomness, and Uncertainty 4.3 Basic Counting Rules In this problem, order is important. We have n = 23 and r = 3. Solution:

Consider that a cafeteria is serving the following vegetables for lunch one day: carrots, broccoli, spinach, baked beans, corn, and green beans. Suppose you will to order a vegetable plate with 4 different vegetables. How many ways can this plate be prepared? Answer the following: HAWKES LEARNING SYSTEMS math courseware specialists Probability, Randomness, and Uncertainty 4.3 Basic Counting Rules In this problem, order is not important. We have n = 6 and r = 4. Solution:

The number of permutations of n objects in which k 1 are all alike, k 2 are all alike, etc., is given by Probability, Randomness, and Uncertainty 4.3 Basic Counting Rules HAWKES LEARNING SYSTEMS math courseware specialists Special Permutations:

How many different ways can you arrange the letters in the word Tennessee? Answer the following: HAWKES LEARNING SYSTEMS math courseware specialists Probability, Randomness, and Uncertainty 4.3 Basic Counting Rules We cannot make a distinction between each “e”, “n”, or “s”. We have T = 1, E = 4, N = 2, S = 2. Solution: