Respuesta :
There are 715 ways to select 9 players for the starting lineup.
Permutation is used whenever there is arrangement or where order is important . denoted by [tex]P(n,r)=\frac{n!}{(n-r)!}[/tex]
n= no of item , r = no of items to be arranged.
Combination is used when there is selection . denoted by[tex]C(n,r)=\frac{n!}{r!(n-r)!}[/tex]
n=no of item , r= no of item to be selected.
Part (a)
In the given question
we have to select 9 players out of 13 player , combination will be used
[tex]C(13,9)=\frac{13!}{9!*(13-9)!}=\frac{13*12*11*10*9!}{9!*4!} =\frac{13*12*11*10}{4*3*2}[/tex]=715 ways.
Part(b)
In this part we have to find how many ways are there to select 9 players for the starting lineup and a batting order for the 9 starters.
Since the batting order is important , permutation will be used.
[tex]P(13,9)=\frac{13!}{4!} =\frac{13*12*11*10*9*8*7*6*5*4!}{4!} =13*12*11*10*9*8*7*6*5[/tex]
=259459200 ways.
Part(c)
In this part we have to do selection of 3 left-handed outfielders and have all other 6 positions occupied by right-handed players.
Since order is not important Combination will be used.
To select 3 left handers from total 6 left handers = C(6,3)
& to select 6 positions of left handers from remaining 7 right handers=C(7,6)
No of ways of selection = C(6,3)*C(7,6)
[tex]=\frac{6!}{3!*(6-3)!} *\frac{7!}{6!*(7-6)!} \\ \\= \frac{6!}{3!*3!} *\frac{7!}{6!*1!} \\ \\[/tex]
On solving further we get
[tex]= \frac{7!}{3!*3!} =\frac{7*6*5*4*3!}{3!*3*2*1} =7*5*4=140ways[/tex]
Therefore ,(a)There are 715 ways to select 9 players for the starting lineup.
(b) there are 259459200 ways to to select 9 players for the starting lineup and a batting order for the 9 starters and 140 ways .
(c)there are 140 ways to select 3 left-handed outfielders and have all other 6 positions occupied by right-handed players.
Learn more about Permutation& Combination here https://brainly.com/question/13480867
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