Question

For this problem, assume 7 males audition, one of them being George, 6 females audition, one of them being Margaret, and 4 children audition. The casting director has 3 male roles available, 1 female role available, and 2 child roles available. (1) How many different ways can these roles be filled from these auditioners?

(2) How many different ways can these roles be filled if exactly one of George and Margaret gets a part?
(3) What is the probability (if the roles are filled at random) of both George and Margaret getting a part?

Answer 1

1.1296

2.570

3.
`(6)/(91)`

Explanation:

We are given that

Number of males=7 including George

Number of females=6 including Margaret

Number of children=4

Number of male selecting for roles=3

Number of females selecting for roles=1

Number of child selecting for roles=2

1.We have to find the number of ways can these roles be filled from these auditioners.

Total number of ways=
`7C_2* 6C_1* 4C_2`

`nC_r=(n!)/(r!(n-r)!)`

Using this formula

Total number of ways=
`(9* 8* 7!)/(2!7!)* (6* 5!)/(1!5!)* (4* 3* 2!)/(2!* 2* 1)`

Total number of ways=1296

2.We have to find number of ways can these roles be filled if exactly one of George and Margaret gets a part.

If George gets a part then Margaret out

Total number of ways=
`6C_2* 5C_1* 4C_2=(6* 5* 4!)/(2* 1\cdot4!)* 5* (4* 3* 2!)/(2* 1\cdot 2!)=450`

If Margaret gets a part then George out

Number of ways=
`6C_3* 4C_2=(6* 5* 4* 3!)/(3* 2* 1* 3!)* (4* 3* 2)/(2!* 2* 1)=120`

Therefore, total number of ways can these roles be filled if exactly one of George and Margaret gets a part=450+120=570

3.We have to find the probability of both George and Margaret getting a part.

Total number of audition=7+6+4=17

Except George and Margaret , number of auditions=15

Number of males=6

Probability=
`(number\;of\;favorable\;cases)/(total\;number\;cases)`

The probability of both George and Margaret getting a part=
`(6C_2* 4C_2)/(15C_4)=((6!)/(2!4!)* (4!)/(2!2!))/((15!)/(4!11!))`

The probability of both George and Margaret getting a part=
`(6* 5* 4!)/(2* 1* 4!)* (4* 3* 2!)/(2!* 2* 1)* (4* 3* 2* 11!)/(15* 14* 13* 12* 11!)`

The probability of both George and Margaret getting a part=
`(6)/(91)`