Question

Jaimie tells 10 of his friends that he just won a vacation to Disneyworld. Unfortunately, he then finds out he can only take 3 of them. He decides to let the other 7 down easy by telling them the large number of ways he could've picked 3 of them to take. Would this be a combination or permutation problem? Why?

Answer 1

Jaimie's problem of picking 3 friends out of 10 to go to Disneyworld is a combination problem because the order of selection does not matter. The number of ways to choose 3 friends from 10 is calculated using the formula 10C3, which results in 120 different combinations.

Step-by-step explanation:

When Jaimie tells 7 of his friends about the large number of ways he could have picked to take 3 of them to Disneyworld, he is referring to a combination problem, not a permutation. The difference between combinations and permutations is important: combinations are used when the order of selection doesn't matter, while permutations are used when the order does matter. Since the order in which the friends are chosen to go to Disneyworld is irrelevant, we use combinations to calculate the number of ways Jaimie could choose 3 friends out of 10.

To calculate the number of combinations, we use the formula for combinations which is nCr = n!/(r! (n-r)!), where n is the total number of items to choose from, r is the number of items to choose, and '!' denotes factorial. So in this case, we calculate 10C3 = 10!/(3!(10-3)!) = 10!/(3!7!) = (10*9*8)/(3*2*1) = 120. Therefore, there are 120 different combinations of friends that Jaimie could invite to Disneyworld.