Chapter_4.4

Slide 1:

© Rita Marie O’Brien Chapter 4.4 Page 1 CHAPTER 4 - 4 Counting Rules This section is asking for possible ways things can be arranged. It is not asking for the probability.

Slide 2:

© Rita Marie O’Brien Chapter 4.4 Page 2 CHAPTER 4 - 4 Factorial Formulas n! = n(n-1)(n-2)…1 0! = 1 Example:

Permutations:

© Rita Marie O’Brien Chapter 4.4 Page 3 Permutations A permutation is an arrangement of n objects in a specific order . The arrangement of n objects in a specific order using r objects at a time is called a permutation of n objects taking r objects at a time. . It is written as n P r , and the formula is:

Slide 4:

© Rita Marie O’Brien Chapter 4.4 Page 4 Permutations n P r Is read n permute r. The n is the total number of choices and the r is the amount you want to choose. Therefore, n is always the larger number. Remember the most important fact of permutations is that order matters.

Slide 5:

© Rita Marie O’Brien Chapter 4.4 Page 5 Example 8 P 2 = 7 P 5 = 12 P 4 = 5 P 3 =

Fill in the permutations before continuing.:

© Rita Marie O’Brien Chapter 4.4 Page 6 Fill in the permutations before continuing.

Slide 7:

© Rita Marie O’Brien Chapter 4.4 Page 7 Example 8 P 2 = 56 7 P 5 = 2,520 12 P 4 = 11,880 5 P 3 = 60

Slide 8:

© Rita Marie O’Brien Chapter 4.4 Page 8 Problem 2 9 P 9 = 362,880 or

Slide 9:

© Rita Marie O’Brien Chapter 4.4 Page 9 Problem 8 Call letters K_ _ _ or W_ _ _ The first choice is a K or W (2 choices) a) If repetitions are allowed for the second letter you have 26 choices (26 letters of Alphabet), there are 26 choices for both the 3 rd and 4 th letters. Ans. 2(26)(26)(26)=35,152

Slide 10:

© Rita Marie O’Brien Chapter 4.4 Page 10 Problem 8 Call letters K_ _ _ or W_ _ _ The first choice is a K or W ( 2 choices) b) If no repetitions: 2 nd letter there are 25 choices (one for the 1st letter) 3 rd letter 24 choices (one for the 1st letter and one for the 2 nd ) 4 th letter 23 choices. Ans. 2(25)(24)(23)= 27,600

Slide 11:

© Rita Marie O’Brien Chapter 4.4 Page 11 Problem 14 Order matters, therefore, it is a permutation. There are 8 homes from which to choose, so n=8 There are to be 8 to be assessed, so r = 8 8 P 8 = 40,320 (NOTE: Since no repetitions = 8! = 40,320) There are 5040 ways to arrange the soap.

Slide 12:

© Rita Marie O’Brien Chapter 4.4 Page 12 Problem A In a board of directors composed of 8 people, how many ways can 1 chief executive officier, 1 director, and 1 treasurer be selected?

Slide 13:

© Rita Marie O’Brien Chapter 4.4 Page 13 Problem A Order matters, so it is a permutation. There are 8 people from whom to choose, so n=8 There are to be 3 people chosen with a rank, so r = 3 8 P 3 = 336 There are 336 ways to choose the 3 officers from a group of 8 people.

Slide 14:

© Rita Marie O’Brien Chapter 4.4 Page 14 Problem 56 How may different signals can be made using at least three distinct flags if there are five different flags from which to select?

Slide 15:

© Rita Marie O’Brien Chapter 4.4 Page 15 Problem 56 Order matters, so it is a permutation. There are five flags, so n=5 The words at least 3 means there could be 3 or 4 or 5 so, r = 3, r = 4 and r = 5 That makes 3 different outcomes and we wish to know how many ways all 3 can happen, so we need to add all the ways. In other words, the ways to use 3 plus the ways to use 4 plus the ways to use 5. 5 P 3 + 5 P 4 + 5 P 5 = 300 There are 300 ways to choose at least 3 flags.

Combinations:

© Rita Marie O’Brien Chapter 4.4 Page 16 Combinations A selection of distinct objects without regard to order is called a combination . The number of combinations of r objects selected from n objects is denoted n C r and is given by the formula:

Slide 17:

© Rita Marie O’Brien Chapter 4.4 Page 17 Combinations Is read n choose r. The n is the total number of choices and the r is the amount you want to choose. Therefore, n is always the larger number. Remember the most important fact of combinations is that order does not matter .

Slide 18:

© Rita Marie O’Brien Chapter 4.4 Page 18 Combinations 3 C 2 = 7 C 3 = 9 C 3 =

Fill in the Combinations before continuing.:

© Rita Marie O’Brien Chapter 4.4 Page 19 Fill in the Combinations before continuing.

Slide 20:

© Rita Marie O’Brien Chapter 4.4 Page 20 Combinations 3 C 2 = 3 7 C 3 = 35 9 C 3 = 84

Slide 21:

© Rita Marie O’Brien Chapter 4.4 Page 21 Problem 28 Order does not matter, so it is a combination. There are 52 cards from which to choose, so n=52 The are to be 3 cards so, r = 3 There are 22,100 ways to choose 3 cards from a deck of 52 cards.

Slide 22:

© Rita Marie O’Brien Chapter 4.4 Page 22 Problem 36 Order does not matter, so it is a combination. The are 23 vehicles but, you can not choose from the group of different cars, therefore there will be three combinations. One has 4 tank cars in which to chose 2, an other has 12 boxcars in which to choose 5, and the third has 7 flatcars in which to chose 3. The multiplication rule then applies.

Note:

© Rita Marie O’Brien Chapter 4.4 Page 23 Note If there are two or more groups from which to choose, then the multiplication rule applies. If phrases such as those listed below are part of a problem, then you will need to add the permutations or combinations. At least At most Less than More than

Conclusions:

© Rita Marie O’Brien Chapter 4.4 Page 24 Conclusions A tree diagram can be used when a list of all possible outcomes is necessary. When only the total number of outcomes is needed, the multiplication rule, the permutation rule, and the combination rule can be used.