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A group of 5 friends—Archie, Betty, Jerry, Moose, and

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

A. 32
B. 36
C. 48
D. 72
E. 120
[Reveal] Spoiler: OA
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Re: A group of 5 friends—Archie, Betty, Jerry, Moose, and [#permalink] New post 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.
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Re: A group of 5 friends—Archie, Betty, Jerry, Moose, and [#permalink] New post 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.
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Re: A group of 5 friends—Archie, Betty, Jerry, Moose, and [#permalink] New post 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
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Re: A group of 5 friends—Archie, Betty, Jerry, Moose, and [#permalink] New post 23 Dec 2012, 13:44
A,J, M & B,V

AISLE SEAT X THIRD ROW X FIRST ROW

Case 1: B/V sit in third row

3 * 2 * 3*2*1 = 36

Case 2: B/V sit in first row

3 * 2 * 2*1*1 = 12

Total ways = 36 +12 = 48

Hence C
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Re: A group of 5 friends—Archie, Betty, Jerry, Moose, and [#permalink] New post 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

Hope this helps....
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Re: A group of 5 friends—Archie, Betty, Jerry, Moose, and [#permalink] New post 13 Jan 2013, 02:23
Hey...I am not able to understand the explanations given
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Re: A group of 5 friends—Archie, Betty, Jerry, Moose, and [#permalink] New post 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

Hence required probability = 72-24 = 48
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Re: A group of 5 friends—Archie, Betty, Jerry, Moose, and [#permalink] New post 09 Aug 2013, 06:00
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006 wrote:
Hey...I am not able to understand the explanations given


solution: use this diagram
A for Archie J for Jerry etc here
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aisle.png
aisle.png [ 38.62 KiB | Viewed 1184 times ]


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Re: A group of 5 friends—Archie, Betty, Jerry, Moose, and [#permalink] New post 09 Aug 2013, 06:03
Asifpirlo wrote:
006 wrote:
Hey...I am not able to understand the explanations given


solution: use this diagram
A for Archie J for Jerry etc here

go stepwise ... B or V in the 3rd row (forgot to mention V in the top right of diagram)
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Re: A group of 5 friends—Archie, Betty, Jerry, Moose, and   [#permalink] 09 Aug 2013, 06:03
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