Q:

# There are 20 members of a basketball team.(a) The coach must select 12 players to travel to an away game. How many ways are there to select the players who will travel?(b) From the 12 players who will travel, the coach must select her starting line-up. She will select a player for each of the five positions: center, right forward, left forward, right guard, left guard. How many ways are there for her to select the starting line-up?(c) From the 12 players who will travel, the coach must select her starting line-up. She will select a player for each of the five positions: center, right forward, left forward, right guard, left guard. However, there are only three of the 12 players who can play center. Otherwise, there are no restrictions. How many ways are there for her to select the starting line-up?

Accepted Solution

A:
Answer:a)125970b)95040c)35640Step-by-step explanation:a) selecting 12 player from 20 players can be one in $$^{20} C_{12}$$  $$^{20} C_{12}$$ = $$\frac{20!}{12!X (20-12)!}$$           =  $$\frac{20!}{12!X8!}$$           = $$125970$$b) since each position has an identity permutations can be used so    Number of ways = $$^{12} P_{5}$$                                 = $$\frac{12!}{(12-5)!}$$                                 = $$\frac{12!}{7!}$$                                 = $$95040$$c) coach permutes 4 players for position other than centers in  $$^{12}P_{4}$$ ways and selects center in 3 ways   so, number of ways = $$^{12}P_{4}X3$$                                      =$$\frac{12!}{8!}X3$$                                      =$$35640$$