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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]

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29 Sep 2012, 13:42
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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]

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29 Sep 2012, 15:11
3
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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.

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

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07 Dec 2012, 19:19
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.

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]

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08 Dec 2012, 05:19
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]

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23 Dec 2012, 14:44
1
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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]

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24 Dec 2012, 09:38
2
KUDOS
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.

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]

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13 Jan 2013, 03:23
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]

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09 Aug 2013, 05:16
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]

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09 Aug 2013, 07:00
2
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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
Attachments

aisle.png [ 38.62 KiB | Viewed 5160 times ]

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

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09 Aug 2013, 07:03
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]

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

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08 Mar 2015, 09:21
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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, so 3 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, so 4 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

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

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07 Jul 2015, 20:53
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.

4 3 2 "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):

B V _ "3"
_

V B _ "3"
_

_ B V "3"
_

_ V B "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
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Re: A group of 5 friends—Archie, Betty, Jerry, Moose, and [#permalink]

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25 Aug 2015, 23:30
I know this is an old post but still -
This is how I would have solved –
I have 5 seats with one being aisle and 5 individuals.
Total number of ways these 5 persons can be seated without any restriction is 5! = 120

Now looking The restrictions
First restriction - Archie, Jerry, or Moose must sit in the aisle seat.
How many different ways can these three not sit in the aisle seat?
In this case I have only 2 possibilities (Betty and Veronica) for the aisle seat.
Aisle Seat – 2 possibilities
Single seat in the 3rd row – 4 possibilities
Middle 3 seats – 3! = 6 possibilities
Thus, 2(Aisle Seat)*4(Single seat in the 3rd row)*6(middle 3 seats) = 48 ways
Remember these are 48 ways that we are not interested in, so subtract them from 120 – 48 = 72.

Restriction 2 - Betty and Veronica refuse to sit next to each other.
How many different ways can these two actually sit together – this is possible only with the middle three seats, therefore –
Aisle seat – 3 possibilities (all but Betty and Veronica)
Single seat in the 3rd row – 2 possibilities
Middle 3 seats - four different ways we can arrange them (seating Betty and Veronica together)
(BVX, XBV, VBX, XVB)
So this gives us – 3(Aisle Seat)*2(Single seat in the 3rd row)*4(middle 3 seats) = 24.
Again these are possible ways that we don’t want, so subtract them from the remaining 72 possible options that we have- 72 – 24 = 48
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A group of 5 friends—Archie, Betty, Jerry, Moose, and [#permalink]

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26 Aug 2015, 04:23
Aisle Seat 1st seat 2nd-Seat 3rd-Seat Adjacent seat
Max probability 3C1 4 3 2 1

Therefore max probability= 3*4*3*2 = 72

lets assume Betty and veronica sit together
So we have 3c1 (BV) J/A A

Taking BV as one unit So 2!. the third seat can be taken in 2 different ways. Betty and Veronica be arranged in 2 different ways.

Required probability = 3C1 * 2!*2*2 = 24

Therefore req ways = 72- 24= 48
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Re: A group of 5 friends—Archie, Betty, Jerry, Moose, and [#permalink]

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30 Aug 2016, 04:50
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Re: A group of 5 friends—Archie, Betty, Jerry, Moose, and [#permalink]

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17 Sep 2016, 16:25
This is what I did:

Total combinations possible without any restrictions: 5! = 120 unique combinations.

Given restrictions 1) A, J or M has to be in the aisle seat & 2) B and V cannot be seated together. (Why can't the guys in these problems just get along? )

Let's try to break restrictions.

To break restriction 1, these are two cases.
If B takes the aisle seat, the rest can be arranged in 4! ways = 24 combinations.
If V takes the aisle seat, the rest can be arranged in 4! ways = 24 combinations.
So total combinations that break this restriction is 24 + 24 = 48.

To break restriction 2:
I visualized the scenario as below.

(Aisle Seat) _ _ _ (Middle seats)

_(3rd row seat)

The only way B and V will be together is if there occupy the middle seats. Think of just one below example:

A BVJ

M

In the case above BVJ can be organized in 2 ways: BVJ or JBV. Since we need to consider VB seating as an option, VBJ and JVB are also possible. So we have 4 options.

Regardless of which person takes up the third seat, there will always be 4 combinations. The Aisle, third middle seat and 3rd row seat can be arranged in 3! ways. So 6 ways. For each of these 6 ways, there are 4 combinations of middle row seating with B & V beside each other.

Finally total cases: 6*4=24

Finally total combinations considering restrictions 1 and 2 = (Total combinations without any restrictions) - (Total combinations which break restriction 1) - (Total combinations which break restriction 2) = 120 - 48 - 24 = 48.

Kudos if you like my explanation
Re: A group of 5 friends—Archie, Betty, Jerry, Moose, and   [#permalink] 17 Sep 2016, 16:25
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