Question

Rework problem 35 from the Chapter 2 review exercises in your text, involving auditioning for a play. 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.

Answer 1

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