Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

A group of 5 friends—Archie, Betty, Jerry, Moose, and [#permalink]
29 Sep 2012, 12:42

1

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

75% (hard)

Question Stats:

47% (03:06) correct
53% (01:24) wrong based on 74 sessions

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?

Re: A group of 5 friends—Archie, Betty, Jerry, Moose, and [#permalink]
29 Sep 2012, 14:11

smartass666 wrote:

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?

32 36 48 72 120

First, place one person in the isle seat - 3 possibilities. Second, place the two girls Betty and Veronica (more specifically separate them) and then place the remaining two persons - the girls can sit either in the middle seats, one seat apart, or one girl in the third row and then the other girl with the remaining two guys in the middle seats - which is 2*2! + 2*3! = 16.

Total number of possibilities 3*16 = 48.

Answer C. _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: A group of 5 friends—Archie, Betty, Jerry, Moose, and [#permalink]
07 Dec 2012, 18:19

EvaJager wrote:

smartass666 wrote:

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?

32 36 48 72 120

First, place one person in the isle seat - 3 possibilities. Second, place the two girls Betty and Veronica (more specifically separate them) and then place the remaining two persons - the girls can sit either in the middle seats, one seat apart, or one girl in the third row and then the other girl with the remaining two guys in the middle seats - which is 2*2! + 2*3! = 16.

Total number of possibilities 3*16 = 48.

Answer C.

Could you please elaborate on this further. I am still stuck with the question.

Re: A group of 5 friends—Archie, Betty, Jerry, Moose, and [#permalink]
08 Dec 2012, 04:19

smartass666 wrote:

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?

32 36 48 72 120

I also struggle with probability and combinations Need some more input from some one _________________

GMAT - Practice, Patience, Persistence Kudos if u like

Re: A group of 5 friends—Archie, Betty, Jerry, Moose, and [#permalink]
24 Dec 2012, 08:38

aditi2013 wrote:

EvaJager wrote:

smartass666 wrote:

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?

32 36 48 72 120

First, place one person in the isle seat - 3 possibilities. Second, place the two girls Betty and Veronica (more specifically separate them) and then place the remaining two persons - the girls can sit either in the middle seats, one seat apart, or one girl in the third row and then the other girl with the remaining two guys in the middle seats - which is 2*2! + 2*3! = 16.

Total number of possibilities 3*16 = 48.

Answer C.

Could you please elaborate on this further. I am still stuck with the question.

Hi,

I used the following method, please confirm this method though as this is one of my weak areas...:-

Consider the following seating arrangement:-

S1, S2 S3, S4 and S5 (aisle)

Now, as per one of the restrictions, S5 can be filled in 3 ways only, and once it is filled, no other arrangement is possible ( as it is one seat)

Therefore, S5 (fill) = 3ways

We are now left with 4 seats and 4 individuals:

Number of ways of selecting 4 individuals to fill 4 seats = 4C4 = 4 ways Moreover, among those 4 individuals, we can have 4! arrangements = 24 arrangements

Therefore, the total arrangements = 24 *4 = 72 arrangements -----This is only when the second restriction is in effect

Lets us now, consider restriction 1: Betty and Veron are never together

No of arrangements when betty and veron are never together = (Total number of arrangements) - (No of arrangements when betty and veron are always together)

When betty and Veron are always together: Consider both one entity, Lets say = X

there fore, now we have 4 seats and 3 people: Number of ways 3 peoplefill 4 seats = 4C3 = 4 ways Number of arrangeents among those 3 people = 3! = 6 arrangements

Therefore, total number of arrangements when Betty and Veron sit together are = 4* 6 = 24 arrangements

=>No of arrangements when betty and veron are never together = 72 - 24 = 48 arrangements

Re: A group of 5 friends—Archie, Betty, Jerry, Moose, and [#permalink]
09 Aug 2013, 04:16

There are 5 seats one in 3rd row, and 4 in first row. It should look something like this *****5- 1-2-3- **4- ; where 1,2,3 are middle seat, 4 is the aisle seat and 5 is the 3rd row seat. Now there are five people, A,B,J,M,V. Now, At least one of A,J,M must sit on seat 4, B,V must not sit together

Lets assume that B,V are together, in that case one of the arrangements would appear something like this *****M B-V-J**A Now looking at the situation that B,V are together and either of A,J,M could sit in seat 4, total no. of possibilities are 2(B-V together and with J)*2(for B,V swapping seats)*3(as at least one of A,J,M must sit in seat 4)*2(for J and M swapping seats) = 24

Now total no. of possibilities with the only restriction that at least one of A,J,M must sit in seat 4 = 4!*3 = 72