Question

A starting lincup in baaketball consizt of two guards, two forwards, and a center A cer ain collage team na前on its mstar our cente s, five uards, Iya n ards, and āna indlvidual X who can play alt ar guard nr n ard. How many n fferent starting nFups can he craate ? Hunt: Con then lineups with X as orward. er I naups without than linAllps with X guard lincups h) Now supposa the rostar has guards,5 forwards, 4 cantars, and 2 swing playars (X and Y) who can play alter guad ar forward. Tf S of tha 15 nlayers ara randomly salertad, what is tha prohability that thay constitta a lagitimate starting lineup? (Raund your answer to three decimal places.)

Answer 1

b)

first consider lineups without X,

the number of line ups without Y ,with Y as guard,with Y as forward given below

[C(4,2)*C(5,2)*C(4,1)+C(4,1)*C(5,2)*C(4,1)+C(4,2)*C(5,1)*C(4,1)=(240+160+120)=520

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now, consider lineups with X as guard,then

the number of line ups without Y ,with Y as guard,with Y as forward given below

[C(4,1)*C(5,2)*C(4,1)+C(4,0)*C(5,2)*C(4,1)+C(4,1)*C(5,1)*C(4,1)=(160+40+80)=280

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now, consider lineups with X as forward,then

the number of line ups without Y ,with Y as guard,with Y as forward given below

[C(4,2)*C(5,1)*C(4,1)+C(4,1)*C(5,1)*C(4,1)+C(4,2)*C(5,0)*C(4,1)=(120+80+24)=224

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total number of legitimate lineups=520+280+224=1024

5 players are selected out of 15 in C(15,5) ways=3003

P(legitimate line ups) = 1024/3003 = 0.341(answer)