Exercise 149 Page 702

a We want to find the number of ways to select scoops if we choose three different flavors from 23 possible, and the order does matter. We have 23 choices for the first scoop, 23-1=22 choices for the second one, and 23-2 = 21 choices for the last one. The number of ways to select three flavors is the product of these numbers.

23 * 22 * 21 = 10 626 Nora has 10 626 ways to choose the scoops.

b This time, Nora wants a dish, so the order of the scoops does not matter. We want to find the number of ways to select scoops if we choose three different flavors from 23 possible, and the order does not matter. In Part A we found the number of ways to choose these flavors if the order does matter.

10 626ways

Now, let's notice that we have 6 different ways to arrange every choice of three items. ccc Item1 Item2 Item3 && Item1 Item3 Item2 Item2 Item1 Item3 && Item2 Item3 Item1 Item3 Item1 Item2 && Item3 Item2 Item1 This means every choice of three different flavors when the order does not matter gives us 6 ways to choose these flavors when the order does matter. Let's say Nora has x ways to choose the scoops for a dish. Then she has 6x ways to choose scoops when caring about the scoops' order, giving us the following equation. 6x = 10 626 ⇒ x = 1771 Nora has 1771 ways to choose scoops for a dish.

c Connie wants a dark chocolate scoop at the bottom and any other two scoops. There are 23 flavors overall, so she chooses the other two scoops from 23-1=22 left. Since they choose from 22 flavors twice, they have 22*22 = 484 ways to do so.

d Vlad has 4 choices for the chocolate bottom flavor. Then, he chooses from 23-4=19 non-chocolate flavors for the other two scoops. Therefore, he has 19*19 = 361 ways to select the other two scoops. The total number of ways to create such a dessert is the product of the ways to choose the bottom flavor and the other two flavors.

4 * 361 = 1444

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Exercise 149 Page 702

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