Unit 8A

Note that I have used the factorial notation in the solutions that involve permutations and combinations. It is not necessary to use this notation; the methods discussed in class give the same results.

8. You have two choices about air conditioning, two choices about a sunroof, two choices about a CD player, and eight choices about color. There are 2 ´ 2 ´ 2 ´ 8 = 64 different styles of car from which to choose.

10c. There are 26 letters that can be placed (with repetition allowed) in each of the six positions of the word. This means that there are

26 ´ 26 ´ 26 ´ 26 ´ 26 ´ 26= 26⁶ = 308,915,776

different words that can be created.

14d. This is also a permutation problem: five contestants must be chosen from a group of ten people and order does matter. The number of permutations is

There are 30,240 different schedules for the first night (and another 30,240 different schedules for the second night).

16c. The statement that order is not important is the clue that combinations are involved. The number of ways that five disks can be selected from a group of twelve disks is

The twelve disks can be arranged in 792 different groups of five if order does not matter and no repetition is allowed.

18. This problem requires using all four of the counting methods discussed in the text.

a. We use the multiplication principle to count arrangements made from two different groups. If there are 12 different flavors and six different toppings, 12 ´ 6 = 72 different sundaes can be made.

b. If the same flavor can be used more than once, we must count arrangements with repetition. There are 12 ways to choose the first scoop, 12 ways to choose the second scoop, and 12 ways to choose the third scoop. This amounts to 12 ´ 12 ´ 12 = 12³ = 1728 different triple cones.

c. Without repeated flavors, we are counting permutations or combinations. If the order of flavors on the cone matters (as in this problem), then we are counting permutations. In this case, we have 12 flavors which must be chosen three at a time. Therefore, the number of triple cones that can be made from 12 flavors with no repeated flavors is

where the order matters.

d. In this problem we are told that order does not matter (vanilla, chocolate, strawberry is the same as , chocolate, vanilla, strawberry). Thus we are counting combinations. The total number of cones possible is now

As always, once n and r are fixed, the number of combinations is less than the number of permutations.

e. If there are 15 flavors in the shop, then the same arguments lead to the following results: (a) there are 15 ´ 6 = 90 different sundaes, (b) there are 15³ = 3375 different triple cones with repetition, (c) there are ₁₅P₃ = 2730 triple cones where order does matter, and (d) ₁₅C₃ = 455 triple cones where order does not matter.

28. The best way to approach this problem is by trial and error. Assume that Luigi has four different toppings. Then the number of three-topping pizzas he can make is given by the combination formula:

If he has five different toppings, he can make ₅C₃ = 10 pizzas. With six toppings, he can make ₆C₃ = 20 pizzas. With seven toppings, he can make ₇C₃ = 35 pizzas. With eight toppings, he can make ₈C₃ = 56 pizzas. And with nine toppings, he can make ₉C₃ = 84 pizzas. We can conclude that Luigi uses nine toppings. The same kind of reasoning can be applied to Ramona and her two-topping pizzas. If she has five toppings, she can make ₅C₂ = 10 pizzas. If she has six toppings, she can make ₆C₂ = 15 pizzas. If she has seven toppings, she can make ₇C₂ = 21 pizzas. If she has eight toppings, she can make ₈C₂ = 28 pizzas. If she has nine toppings, she can make ₉C₂ = 36 pizzas. If she has ten toppings, she can make ₁₀C₂ = 45 pizzas. Whew! She must use ten toppings.

Unit 8B

12. There are 45 possible outcomes (because there are 45 candies in the bag).

a. A success is drawing a chocolate, of which there are ten. Thus, the probability of drawing a chocolate is 10/45 = 2/9.

b. A success is drawing a mint, of which there are 15. Thus, the probability of drawing a mint is 15/45 = 1/3.

c. A success is drawing a gumdrop, of which there are 20. Thus, the probability of drawing a gumdrop is 20/45 = 4/9.

d. A success is drawing a chocolate or gumdrop (not a mint). There are 30 chocolates or gumdrops. Thus, the probability of not drawing a mint is 30/45 = 2/3, which is one minus the probability of drawing a mint.

14c. Because there are 36 cards with 2 through 10 in a deck of 52 cards, P(2 through 10) = 36/52 = 9/13. Thus P( 2 through 10) = 1 - 9/13 = 4/13.

16. We can refer to the table in Example 5 of the text.

a. Because there are 4 ways that the two dice can have a sum of 5 and there are 36 possible outcomes, P(sum of 5) = 4/36 = 1/9.

b. Using part (a), P(not sum of 5) = 1 - P(sum of 5) = 1 - 1/9 = 8/9.

c. There are two ways to roll a 6 on one die and a 1 on the other die. Therefore P(1 and 6) = 2/36 = 1/18.

d. The table shows that the number of times that one (but not both) of the dice is a 5 is 10. The total number of outcomes is 36, so P(one 5) = 10/36 =5/18.

e. Imagine that you play this game 36 times (although you can pick any number of games). In 36 games you would expect (on average) to roll a sum of 7 and win 6 times. You would expect to roll a sum other than 7 and lose 30 times. Therefore, you would expect to win a total of 6 ´ $10 = $60 and lose a total of 30 ´ $1 = $30. Thus, in the long run, you would expect to win $2 for every $1 you lose. This game is probably worth playing.