At a game show, there are 8 people (including you and your friend) in the front row. The host randomly chooses 3 people from the front row to be contestants. The order in which they are chosen does not matter. How many ways can you and your friend both be chosen?

Answer:A. C(3,8) = 56Step-by-step explanation:Since the order doesn't matter, it's a combination, not a permutation (answer C).To calculate the possible combinations, we use the following formula:[tex]C(n,r) = \frac{n!}{r! (n-r)!}[/tex]In this case we have n = 8 (entire population), and r = 3 (number of chosen people)[tex]C(8,3) = \frac{8!}{3! (8 - 3)!} = 56[/tex]And you have 56 in your possible answers... so answer A.