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

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Re: A group of 5 friends—Archie, Betty, Jerry, Moose, and [#permalink]
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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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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]
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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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 19348 times ]

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

Arrangements for Veronica and Betty
1. [ ] [___] [___] --- [ ]
2. [___] [ ] [___] --- [ ]
3. [___] [___] [ ] --- [ ]
4. [ ] [___] [ ] --- [___]

$$3_{Isle} * 4_{Arrangements} * 2_{Swap Veronica Betty} * 2_{Remaining Friends} = 48$$
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A group of 5 friends—Archie, Betty, Jerry, Moose, and [#permalink]
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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

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

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

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.

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Re: A group of 5 friendsArchie, Betty, Jerry, Moose, and [#permalink]
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Re: A group of 5 friendsArchie, Betty, Jerry, Moose, and [#permalink]
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