Respuesta :
Answer: 1) 6300 ways
2) 2520 ways
3) 0.067
Step-by-step explanation:
For this problem, assume 9 males audition, one of them being Winston, 5 females audition, one of them being Julia, and 6 children audition. The casting director has 3 male roles available, 1 female role available, and 2 child roles available.
How many different ways can these roles be filled from these auditioners?
Available: 9M and 5F and 6C
Cast: 3M and 1F and 2C
As it is not ordered: C₉,₃ * C₅,₁ * C₆,₂
C₉,₃ = 9!/3!.6! = 84
C₅,₁ = 5!/1!.4! = 5
C₆,₂ = 6!/2!.4! = 15
C₉,₃ * C₅,₁ * C₆,₂ = 84.5.15 = 6300
How many different ways can these roles be filled if exactly one of Winston and Julia gets a part?
2 options Winston gets or Julia gets it:
1) Winston gets it but Julia no:
8 male for 2 spots
4 females for 1 spot
6 children for 2 spots
C₈,₂ * C₄,₁ * C₆,₂
C₈,₂ = = 8!/2!.6! = 28
C₄,₁ = 4!/1!.3! = 4
C₆,₂ = 6!/2!.4! = 15
C₈,₂ * C₄,₁ * C₆,₂ = 28.4.15 = 1680
2) Julia gets it but Winston does not
8 male for 3 spots
1 female for 1 spot
6 children for 2 spots
C₈,₃ * C₆,₂
C₈,₃ = 8!/3!.5! = 56
C₆,₂ = 6!/2!.4! = 15
C₉,₃ * C₆,₂ = 56.15 = 840
1) or 2) = 1) + 2) = 1680 + 840 = 2520
What is the probability (if the roles are filled at random) of both Winston and Julia getting a part?
8 male for 2 spots
1 female for 1 spot
6 children for 2 spots
C₈,₂ * C₆,₂
C₈,₂ = = 8!/2!.6! = 28
C₆,₂ = 6!/2!.4! = 15
C₈,₂ * C₆,₂ = 28.15 = 420
p = 420/6300 = 0.067
