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

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Difficulty:

85% (hard)

Question Stats:

52% (02:56) correct
48% (01:33) wrong based on 134 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

2

This post received KUDOS

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

2

This post received KUDOS

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

Re: A group of 5 friends—Archie, Betty, Jerry, Moose, and [#permalink]
18 Jan 2015, 11:12

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A group of 5 friends—Archie, Betty, Jerry, Moose, and [#permalink]
08 Mar 2015, 08:21

1

This post was BOOKMARKED

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?

A. 32 B. 36 C. 48 D. 72 E. 120

A,B,J,M and V

conditions: (1) A , J or M must sit in the aisle seat (+ve condition) (2) V and B cannot sit together (-ve condition) [+ve condition must be fulfilled first] (aisle seat)*(front row middle seats)*(third row seat)

(__)*(__ *__ *__)*(__)

(3)*(4*2*2)*(1) =48

EXPLAINATION

First on the aisle seat A,J or M can sit, so3 options [now suppose A sits on the aisle seat](4 people left)

Now on the 1st seat of the middle seats on the front row any of the 4 people can sit, so4 options[now suppose V sits on the first seat](3 people are left)

for the second seat we have 3 people left but B cannot sit with V and as we have placed V on the first seat so 2 options

for the third seat we have 2 people left, so 2 options

and for the seat in the third row only one person is left, so 1 option

Re: A group of 5 friends—Archie, Betty, Jerry, Moose, and [#permalink]
07 Jul 2015, 19:53

Here is how I solved it --

Visualizing the available seating: _ _ _ "_" _ middle seat ( _ _ _ ), aisle seat (which is to the right of the middle seat in quotes "_"), and 3rd row seat (which is off in the separate row).

Strategy is to solve for total number of possibilities given the aisle constraint that Archie, Jay, and Moose must sit in the aisle, and subtract the total number of possibilities with the aisle constraint AND Betty and Veronica sitting next to each other.

1. Solve for total number of possibilities given the aisle constraint that Archie, Jay, and Moose must sit in the aisle

Place 3 in "aisle" spot and solve for total possibilities.

_ _ _ "3"

_

Fill remaining seats with remaining number. Remember first we're solving for all possibility with only the aisle constraint.

432 "3" 1

Multiple these and you get 3*4*3*2*1=72 possibilities given only the aisle constraint.

2. Next solve for number of ways Betty and Veronica DO sit next to each other, still keeping the aisle constraint.

_ _ _ "3" _

4 ways that Betty and Veronica can sit next to each other (see below):

BV _ "3" _

VB _ "3" _

_ BV "3" _

_ VB "3" _

So we have 4 ways Betty and Veronica sit next to each other, and since they occupy two seats, there are only 2 people left to fill the the 2 remaining seats with no constraint, so 4*2*1. So total possible ways of Betty and Veronica sitting next to each other, given the aisle constraint: 4*2*1*3(the aisle constraint)=24

3. possible seating arrangements given only the aisle constraint - possible seating arrangements given aisle constraint and Betty and Veronica sitting next to each other. 72-24 = 28

gmatclubot

Re: A group of 5 friends—Archie, Betty, Jerry, Moose, and
[#permalink]
07 Jul 2015, 19:53

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