Amy is packing her bags for her trip that she won to go on a boat cruise. Amy won 4 tickets to the boat cruise and Amy can only bring 3 people along with her on this trip, but 15 of her friends want to go. How many different types of groups of friends can Amy bring with her?

3x15=45

Am I correct with answer to be 45 choices?

In math there is a concept called a combinatorics. [On your calculator, you should see an nCr button to compute the answer accordingly...]

Since order does not matter with regard to picking Amy's friends, the amount of groups that Amy can bring with her can be calculated by 15C3...

15 represents the number of friends she can CHOOSE from.

3 represents the amount of people she will choose from.

So, with this in mind, there are 455 different combinations in which Amy can choose her friends to go with her!

how many ways can 3 people be selected from a group of 15

this is a combination (order not important)
... the notation is ... 15C3

the calculation is ... 15! / [(15 - 3)! * 3!]
... (15 * 14 * 13) / (3 * 2 * 1)

I appreciate your effort in solving the problem, but the answer is not correct. Let's go through the problem and find the correct answer step by step.

Amy won 4 tickets to the boat cruise, and she can only bring 3 people with her. However, she has 15 friends who want to go on the trip.

To find the number of different groups of friends Amy can bring with her, we need to use combinations. In this case, we want to choose 3 friends from a group of 15.

To calculate the number of combinations, we use the formula:

nCr = n! / (r! * (n - r)!)

Where n is the total number of friends (15) and r is the number of friends Amy can bring (3).

So, let's calculate:

15C3 = 15! / (3! * (15 - 3)!)

Simplifying the equation:

15! / (3! * 12!) = (15 * 14 * 13) / (3 * 2 * 1) = 455

Therefore, there are 455 different groups of friends that Amy can bring with her on the boat cruise.