A group of four girls (Alice, Beth, Cara, and Denise) and four boys (Evan, Francis, George,and Harry) decide to split into two teams to play a game of badminton, Alice and Evan are elected team captains, and they take turns picking players for their teams.

(a) In how many different orders can the remaining children be chosen to be on a team?

(b) What is the probability that Francis is picked last? Express the probability as a decimal. If necessary, round to the nearest hundredth.

Guest Feb 20, 2021

#1**0 **

a. Because there are two team captains, there are six people possibly chosen, and there are 6 people the first captain can pick, 5 people the second, 4 the first again...etc, so there are 6*5*4*3*2 ways or 6!=720 ways.

b. if francis is picked last, that means that there are 5 people to choose from for the first captain because francis cannot be chosen, then 4 for the second, etc, so we have 5*4*3*2 or 5!=120 ways that francis can be picked last, and we divide that by 720 because that's the total amount of ways, and we get 1/6 or 0.17(rounded)

SparklingWater2 Feb 26, 2021