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A family consisting of one mother, one father,

by Vincen » Tue Nov 28, 2017 6:22 am
A family consisting of one mother, one father, two daughters and a son is taking a road trip in a sedan. The sedan has two front seats and three back seats. If one of the parents must drive and the two daughters refuse to sit next to each other, how many possible seating arrangements are there?

A. 28
B. 32
C. 48
D. 60
E. 120

The OA is B .

I don't know how to solve this PS question. Can any expert help me a little? Thanks.
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by GMATGuruNY » Tue Nov 28, 2017 6:26 am
A family consisting of one mother, one father, two daughters and a son is taking a road trip in a sedan. The sedan has two front seats and three back seats. If one of the parents must drive and the two daughters refuse to sit next to each other, how many possible seating arrangements are there?
28
32
48
60
120
Case 1: One daughter in the front passenger seat:
Since one of the 2 parents must drive, number of choices for the driver's seat = 2.
Since either of the 2 daughters can be in the front passenger seat, number of choices = 2.
Number of ways to arrange the 3 remaining people = 3! = 6.
To combine these options, we multiply:
2*2*6 = 24.

Case 2: The two daughters separated in the back seat:
Since one of the 2 parents must drive, number of choices for the driver's seat = 2.
Since either of the 2 daughters can be in the leftmost back seat, number of choices = 2.
Since the 1 remaining daughter must be in the rightmost back seat, number of choices = 1.
Number of ways to arrange the 2 remaining people = 2! = 2.
To combine these options, we multiply:
2*2*1*2 = 8.

Total possible arrangements:
Case 1 + Case 2 = 24+8 = 32.

The correct answer is B.
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by Brent@GMATPrepNow » Tue Nov 28, 2017 9:39 am
Vincen wrote:A family consisting of one mother, one father, two daughters and a son is taking a road trip in a sedan. The sedan has two front seats and three back seats. If one of the parents must drive and the two daughters refuse to sit next to each other, how many possible seating arrangements are there?

A. 28
B. 32
C. 48
D. 60
E. 120
Another, slightly longer, approach:

The restriction about the sisters is somewhat problematic, so let's IGNORE the rule and seat all 5 people without obeying that restriction.

Then once I determine the total number of arrangements, I'll subtract the number of arrangements where the sisters are sitting together.

Number the seats as follows:
Seat #1: driver's seat
Seat #2: passenger's seat
Seats #3, 4, 5: back seats

# of arrangements where we ignore rule about the sisters not sitting together
Take the task of seating all 5 people and break into stages.
Stage 1: Seat someone in seat #1
Only a parent can sit here. So, this stage can be accomplished in 2 ways.
Stage 2: Seat someone in seat #2
Once we have seated someone in seat #1, there are 4 people remaining. So, this stage can be accomplished in 4 ways.
Stage 3: Seat someone in seat #3
At this point, we have already seated 2 people, so there are now 3 people remaining.
So, this stage can be accomplished in 3 ways.
Stage 4: Seat someone in seat #4
There are 2 people remaining, so this stage can be accomplished in 2 ways.
Stage 5: Seat someone in seat $5
This stage can be accomplished in 1 way

By the Fundamental Counting Principle (FCP) we can complete all 5 stages (and thus seat all 5 people) in (2)(4)(3)(2)(1) ways
48 ways

So there are 48 different ways to seat the family such that a parent drives. At this point, the 48 different arrangements include arrangements where the sisters are seated together. So, we need to SUBTRACT the number of arrangements where the sisters are seated together.

There are two cases where the sisters are together.
case 1: the sisters are in seats #3 and #4
case 2: the sisters are in seats #4 and #5

case 1: the sisters are in seats #3 and #4
Once again, we'll take the task of seating everyone and break it into stages:
Stage 1: seat a parent in seat #1.
Must be 1 of 2 parents. So, this stage can be accomplished in 2 ways.
Stage 2: seat a sister in seat #3
Must be 1 of 2 sisters. So, this stage can be accomplished in 2 ways.
Stage 3: seat the other sister in seat #4.
Once we have seated a sister in seat #3, only 1 sister remains. So, this stage can be accomplished in 1 way.
Stage 4: seat someone in seat #2.
At this point, we have seated 3 people, so only 2 people remain. So, this stage can be accomplished in 2 ways.
Stage 5: seat someone in seat #5.
One person remaining. So, this stage can be accomplished in 1 way.

By the Fundamental Counting Principle (FCP) we can complete all 5 stages (and thus seat the sisters are in seats #3 and #4) in (2)(2)(1)(2)(1) ways
8 ways

case 2: the sisters are in seats #4 and #5
We can follow the same steps as above to get 8 more arrangements

So the final answer is 48 - 8 - 8 = [spoiler]32 = B[/spoiler]

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Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. For more information about the FCP, watch our free video: https://www.gmatprepnow.com/module/gmat-counting?id=775

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by [email protected] » Tue Nov 28, 2017 11:16 am
Hi Vincen,

These types of questions can be approached a couple of different ways. There's a "visual" aspect to this question that can help you to take advantage of some shortcuts built into the prompt, so I'm going to use a bit of "brute force" and some pictures to answer this question. Since we're arranging people in seats, we'll end up doing some "permutation math."

M = Mother
F = Father
D1 = 1st Daughter
D2 = 2nd Daughter
S = Son

Front Back
_ _ _ _ _
1st spot = driver

We have 2 restrictions that we have to follow:
1) Either the Father or Mother must be the driver
2) The two daughters CANNOT sit next to one another

Let's put the Mother in the driver's seat and count up the possibilities:

M F (2)(1)(1) Here, the two daughters have to be separated by the son, but either daughter could be in the "first back seat" = 2 options

M D1 (3)(2)(1) Here, with the first daughter up front, the remaining 3 people (F, D2 and S) can be in any of the back seats = 6 options
M D2 (3)(2)(1) Here, we have the same situation, but with the second daughter up front... = 6 options
M S (2)(1)(1) Here, with the son up front, we have the same scenario as we had when the Father was up front = 2 options

Total options with Mother driving = 2+6+6+2 = 16 options

Now we can take advantage of the shortcut I mentioned earlier - We can flip-flop the Mother and Father in the above examples. This will gives us another 16 options with the Father driving.

Total options: 16 + 16 = 32 options

Final Answer: B

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A family consisting of one mother, one father,

by Mo2men » Wed Nov 29, 2017 6:38 am
GMATGuruNY wrote:
A family consisting of one mother, one father, two daughters and a son is taking a road trip in a sedan. The sedan has two front seats and three back seats. If one of the parents must drive and the two daughters refuse to sit next to each other, how many possible seating arrangements are there?
28
32
48
60
120
Case 1: One daughter in the front passenger seat:
Since one of the 2 parents must drive, number of choices for the driver's seat = 2.
Since either of the 2 daughters can be in the front passenger seat, number of choices = 2.
Number of ways to arrange the 3 remaining people = 3! = 6.
To combine these options, we multiply:
2*2*6 = 24.

Case 2: The two daughters separated in the back seat:
Since one of the 2 parents must drive, number of choices for the driver's seat = 2.
Since either of the 2 daughters can be in the leftmost back seat, number of choices = 2.
Since the 1 remaining daughter must be in the rightmost back seat, number of choices = 1.
Number of ways to arrange the 2 remaining people = 2! = 2.
To combine these options, we multiply:
2*2*1*2 = 8.

Total possible arrangements:
Case 1 + Case 2 = 24+8 = 32.

The correct answer is B.
Dear Mitch,
I did it with other logic for case 2.
Since one of the 2 parents must drive, number of choices for the driver's seat = 2.
Since one of either son or one of his parents can sit in from beside driver, number of choices for the driver's seat = 2.
Since daughters are sitting in back seat separated by the other persons, number of choices for the back seat = 2.

Total = 8

Is it correct reasoning?

Thanks
Last edited by Mo2men on Wed Nov 29, 2017 3:27 pm, edited 1 time in total.
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by GMATGuruNY » Wed Nov 29, 2017 2:03 pm
Mo2men wrote:Dear Mitch,
I did it with other logic for case 2.
Since one of the 2 parents must drive, number of choices for the driver's seat = 2.
Since one of either son or one of his parents can sit in from beside driver, number of choices for the driver's seat = 2.
Since daughters are sitting in back seat separated by the other pserons, number of choices for the driver's seat = 2.

Total = 8

Is it correct reasoning?

Thanks
If the step in blue refers not to the driver's seat but to the back seat BEHIND the driver's seat, then this approach is perfect.
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by Mo2men » Wed Nov 29, 2017 3:24 pm
GMATGuruNY wrote:
Mo2men wrote:Dear Mitch,
I did it with other logic for case 2.
Since one of the 2 parents must drive, number of choices for the driver's seat = 2.
Since one of either son or one of his parents can sit in from beside driver, number of choices for the driver's seat = 2.
Since daughters are sitting in back seat separated by the other pserons, number of choices for the driver's seat = 2.

Total = 8

Is it correct reasoning?

Thanks
If the step in blue refers not to the driver's seat but to the back seat BEHIND the driver's seat, then this approach is perfect.
Thanks Mitch. You were right. I copied incorrectly. I will update my post.
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by Scott@TargetTestPrep » Wed Jan 03, 2018 4:04 pm
Vincen wrote:A family consisting of one mother, one father, two daughters and a son is taking a road trip in a sedan. The sedan has two front seats and three back seats. If one of the parents must drive and the two daughters refuse to sit next to each other, how many possible seating arrangements are there?

A. 28
B. 32
C. 48
D. 60
E. 120
We can analyze the problem using the following two cases: 1) the father is the driver and 2) the mother is the driver.

Case 1: The father is the driver.

If the father is the driver, we could have the following subcases:

i) The mother sits in the front row beside the father. Since the daughters refuse to sit next to each other, there are 2 seating arrangements in the back row: Dsd, dsD (D = elder daughter, d = younger daughter, and s = son).

ii) The son sits in the front row beside the father. Since the daughters refuse to sit next to each other, there are 2 seating arrangements in the back row: Dmd, dmD (m = mother).

iii) The elder daughter sits in the front row beside the father. Since, in this case, the two daughters will definitely not sit next to each other, there are 3! = 6 seating arrangements for the 3 remaining people who sit in the back row.

iv) The younger daughter sits in the front row beside the father. Like subcase (iii), there will be 6 seating arrangements in the back row.

As we can see from the above, if the father is the driver, there will be a total of 2 + 2 + 6 + 6 = 16 seating arrangements. We can make the same argument when the mother is the driver. Thus, there will be another 16 seating arrangements, and hence we have a total of 16 + 16 = 32 seating arrangements.

Answer: B

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by Brent@GMATPrepNow » Wed Jan 03, 2018 4:17 pm
A family consisting of one mother, one father, two daughters and a son is taking a road trip in a sedan. The sedan has two front seats and three back seats. If one of the parents must drive and the two daughters refuse to sit next to each other, how many possible seating arrangements are there?

A)28
B)32
C)48
D)60
E)120

OAB
Another, slightly longer, approach:

The restriction about the sisters is somewhat problematic, so let's IGNORE the rule and seat all 5 people without obeying that restriction.

Then once I determine the total number of arrangements, I'll subtract the number of arrangements where the sisters are sitting together.

Number the seats as follows:
Seat #1: driver's seat
Seat #2: passenger's seat
Seats #3, 4, 5: back seats

# of arrangements where we ignore rule about the sisters not sitting together
Take the task of seating all 5 people and break into stages.
Stage 1: Seat someone in seat #1
Only a parent can sit here. So, this stage can be accomplished in 2 ways.
Stage 2: Seat someone in seat #2
Once we have seated someone in seat #1, there are 4 people remaining. So, this stage can be accomplished in 4 ways.
Stage 3: Seat someone in seat #3
At this point, we have already seated 2 people, so there are now 3 people remaining.
So, this stage can be accomplished in 3 ways.
Stage 4: Seat someone in seat #4
There are 2 people remaining, so this stage can be accomplished in 2 ways.
Stage 5: Seat someone in seat $5
This stage can be accomplished in 1 way

By the Fundamental Counting Principle (FCP) we can complete all 5 stages (and thus seat all 5 people) in (2)(4)(3)(2)(1) ways
48 ways

So there are 48 different ways to seat the family such that a parent drives. At this point, the 48 different arrangements include arrangements where the sisters are seated together. So, we need to SUBTRACT the number of arrangements where the sisters are seated together.

There are two cases where the sisters are together.
case 1: the sisters are in seats #3 and #4
case 2: the sisters are in seats #4 and #5

case 1: the sisters are in seats #3 and #4
Once again, we'll take the task of seating everyone and break it into stages:
Stage 1: seat a parent in seat #1.
Must be 1 of 2 parents. So, this stage can be accomplished in 2 ways.
Stage 2: seat a sister in seat #3
Must be 1 of 2 sisters. So, this stage can be accomplished in 2 ways.
Stage 3: seat the other sister in seat #4.
Once we have seated a sister in seat #3, only 1 sister remains. So, this stage can be accomplished in 1 way.
Stage 4: seat someone in seat #2.
At this point, we have seated 3 people, so only 2 people remain. So, this stage can be accomplished in 2 ways.
Stage 5: seat someone in seat #5.
One person remaining. So, this stage can be accomplished in 1 way.

By the Fundamental Counting Principle (FCP) we can complete all 5 stages (and thus seat the sisters are in seats #3 and #4) in (2)(2)(1)(2)(1) ways
8 ways

case 2: the sisters are in seats #4 and #5
We can follow the same steps as above to get 8 more arrangements

So the final answer is 48 - 8 - 8 = [spoiler]32 = B[/spoiler]

-----------------------------------------------------------------------------

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. For more information about the FCP, watch our free video: https://www.gmatprepnow.com/module/gmat-counting?id=775

Then you can try solving the following questions:

EASY
- https://www.beatthegmat.com/what-should- ... 67256.html
- https://www.beatthegmat.com/counting-pro ... 44302.html
- https://www.beatthegmat.com/picking-a-5- ... 73110.html
- https://www.beatthegmat.com/permutation- ... 57412.html
- https://www.beatthegmat.com/simple-one-t270061.html
- https://www.beatthegmat.com/mouse-pellets-t274303.html


MEDIUM
- https://www.beatthegmat.com/combinatoric ... 73194.html
- https://www.beatthegmat.com/arabian-hors ... 50703.html
- https://www.beatthegmat.com/sub-sets-pro ... 73337.html
- https://www.beatthegmat.com/combinatoric ... 73180.html
- https://www.beatthegmat.com/digits-numbers-t270127.html
- https://www.beatthegmat.com/doubt-on-sep ... 71047.html
- https://www.beatthegmat.com/combinatoric ... 67079.html


DIFFICULT
- https://www.beatthegmat.com/wonderful-p- ... 71001.html
- https://www.beatthegmat.com/ps-counting-t273659.html
- https://www.beatthegmat.com/permutation- ... 73915.html
- https://www.beatthegmat.com/please-solve ... 71499.html
- https://www.beatthegmat.com/no-two-ladie ... 75661.html
- https://www.beatthegmat.com/laniera-s-co ... 15764.html

Cheers,
Brent
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