vacationsolz.blogg.se - Permutation of 15 items taken 7 at a time

Should you expect to get the "expected value" in the first game?Ĭ. What would your expected earnings per game be? Round your answer to the nearest penny.ī. Youįind the probabilities of landing on each number and record them in a nice little table seen below.Ī. You quickly calculate the expected value before you decide whether you will play or not. You just finished taking FDMAT108 and remember learning about the expected value. See the picture below of what the spinner looks like. The spinner has four sections labeled 1, 2, 3 and 4 each isĪ different size. Whichever number it lands on, you get that much money back.

You just pay $3 to start the game and spin a big spinner. Just played it and won $10 after playing 3 times. One of your friends comes to you and says that he wants you to play this new game he found. Send me a message if you have any questions about this.You and some of your friends decide to visit the Eastern Idaho state fair. How many ways can you select 4 competitors for the Platinum, Gold, Silver, Bronze Awards out of a field of 20 competitors? (20*19*18*17) How many ways can you select 3 pizza toppings from a set of 7 possible toppings? (7*6*5) How many ways can you select 4 officers from a 12-member club? (12*11*10*9) Other examples of permutations (order matters) without replacement: Where n is the total number of items in the set and r is the number you are going to select. Which can be written in math shorthand as follows: (total number - number to be selected) factorial (total number of items in the whole set) factorial We notice that the solution is in this form:ġ2 * 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 Now let's find out how to use this procedure for every kind of permutation-without-replacement problem. Our solution: 12 * 11 * 10 * 9 = 11,880 possibilities, total How many ways can we put on 12 shirts, 11 hats, 10 pants, and 9 scarves?Īha! That means we can solve this problem the same way: So, this now looks like the familiar counting problem.

There are 12-3 ways we can select the fourth item.

There are 12-2 ways we can select the third item.

There are 12-1 ways we can select the second item.

There are 12 ways we can select the first item.

Then, finally, 4) we will inspect our work and identify a procedure we can use on every problem like this one.įor a permutation (order matters) without replacement (each item can be selected only once), we can notice the following pattern:

2) Because you are selecting 4 items from a set of 12, then it is implied that you can only choose each of those 4 items without replacement. Strategy: 1) Notice that this problem is a permutation because the items can be selected only once, meaning that order matters.