A starting lineup in basketball consists of two guards, two forwards, and a cent

Question

A starting lineup in basketball consists of two guards, two forwards, and a center. A certain college team has on its roster four centers, four guards, five forwards, and one individual (X) who can play either guard or forward. How many different starting lineups can be created? Now suppose the roster has 4 guards, 4 forwards, 4 centers, and 2 "swing players" (X and Y) who can play either guard or forward. If 5 of the 14 players are randomly selected, what is the probability that they constitute a legitimate starting lineup? (Round your answer to three decimal places.)

Explanation / Answer

A) 4 CENTRES, FOUR GUARDS AND 5 FORWARDS

WE HAVE TO CHOOSE 2 GUARDS, 2 FORWARDS AND A CENTRE

NOW THE REQUIRED ANSWER WILL BE 4C2*4C2*5C1 = 180 TOTAL WAYS

B)FIRST WE WILL FIND THAT TOTAL SELECTION OF 5 PLAYERS AMONG 14

= 14C5 = 2002

LEGITIMATE CAN BE FORMED ACCORDING TO LINE UP = TAKING SWING PLAYER IN BOTH CENTRES AND GUARDS

THEREFORE 6C2*6C2*4C1 = 120

PROBABILITY = 120/2002 = 0.059

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