Counting

cypherskull wrote: ↑

Sun Aug 19, 2012 11:33 am

A group of 5 friends—Archie, Betty, Jerry, Moose, and Veronica—arrived at the movie theater to see a movie. Because they arrived late, their only seating option consists of 3 middle seats in the front row, an aisle seat in the front row, and an adjoining seat in the third row. If Archie, Jerry, or Moose must sit in the aisle seat while Betty and Veronica refuse to sit next to each other, how many possible seating arrangements are there?

a.32
b.36
c.48
d.72
e.120

There are two different ways to satisfy the given conditions:
Case i: Betty or Veronica sit in the third row
Case ii: Betty and Veronica sit in the 3 middle seats with someone seated between them

Case i: Betty or Veronica sit in the third row
We can select someone to sit in the third row in 2 ways (either Betty or Veronica will sit here)
We can select someone to sit in the aisle seat in 3 ways (can be Archie, Jerry, or Moose)
There are now 3 people remaining to be seated in the three middle seats in the front row
We can select someone to sit in the leftmost seat in 3 ways (can be any of the 3 remaining people)
We can select someone to sit in the middle seat in 2 ways
We can select someone to sit in the rightmost seat in 1 way
By the Fundamental Counting Principle (FCP), we can complete all 5 steps (and the seat all 5 people) in (2)(3)(3)(2)(1) ways (= 36 ways)

Case ii: Betty and Veronica sit in the 3 middle seats with someone seated between them
We can select someone to sit in the aisle seat in 3 ways (can be Archie, Jerry, or Moose)
We can select someone to sit in the leftmost seat in 2 ways (either Betty or Veronica will sit here)
We can select someone to sit in the rightmost seat in 1 way (will be either Betty or Veronica, depending on who is sitting in the left-most seat)
There are now two people remaining to be seated.
We can select someone to sit in the middle seat in 2 ways
We can select someone to sit in the third row in 1 way
By the FCP, we can complete all 5 steps in (3)(2)(1)(2)(1) ways (= 12 ways)

TOTAL number of ways to seat all five people = 36 + 12 = 48

Answer: C


Cheers,
Brent

cypherskull wrote: ↑

Sun Aug 19, 2012 11:33 am

A group of 5 friends—Archie, Betty, Jerry, Moose, and Veronica—arrived at the movie theater to see a movie. Because they arrived late, their only seating option consists of 3 middle seats in the front row, an aisle seat in the front row, and an adjoining seat in the third row. If Archie, Jerry, or Moose must sit in the aisle seat while Betty and Veronica refuse to sit next to each other, how many possible seating arrangements are there?

a.32
b.36
c.48
d.72
e.120

Solution:

There are three choices for the aisle seat.

Assuming one of Archie, Jerry, or Moose is seated on the aisle seat, there are four remaining people.

If Betty sits in the third row, then the remaining three people can sit in any order in the three seats in the front row. The number of options in this scenario is 3! = 6. Similarly, if Veronica sits in the third row, then again there are 3! = 6 ways to seat the remaining three people in the front row.

If neither Betty nor Veronica sits in the third row, then the seating arrangement for the front row must be B-X-V or V-X-B, where there are two choices for X in each case. Notice that after three people are seated in the front row, there is only one person left, and that person must sit in the third row. Thus, there are 6 + 6 + 2 + 2 = 16 ways to seat the five people that remain after the aisle seat is filled.

Since there are 3 ways to fill the aisle seat and 16 ways to fill the remaining seats, there are 3 x 16 = 48 ways to seat the five friends in total.

Answer: C