## A family consisting of one mother...

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### A family consisting of one mother...

by factor26 » Mon Jan 23, 2012 6:34 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

Source: MGMAT CAT EXAM

I have no idea where to start here!! Help!! Thanks!!

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by pemdas » Mon Jan 23, 2012 7:16 pm
It's simple indeed

Let's consider, Farther is driving first and the pertinent cases

1 daughter sits in the front seat and others seat in the back seats 3!. Two such cases 2*3!
Son is in the front seat, mom is in the middle, 2 ways. Mom is in the front seat and Son is in the middle 2 ways. Total makes 2*3!+2+2=16 ways.

The same number of ways when Mom is driving and Farther takes his turn

16*2=32

b

factor26 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

Source: MGMAT CAT EXAM

I have no idea where to start here!! Help!! Thanks!!
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### A family consisting of one mother...

by rijul007 » Mon Jan 23, 2012 8:05 pm
factor26 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

Source: MGMAT CAT EXAM

I have no idea where to start here!! Help!! Thanks!!
You can start of bu drawing the kind of arrangements

2 in the front
3 at the back
Driver has to be a parent, so we have 2 options for one of the front seats

the two daughters refuse to sit next to each other,
Lets say one of the sisters sit at the front seat
So we have 2 options for the second seat in the front
and 3 people to arrange on 3 seats at the back

Arrangements = 2*2*3*2*1 = 24

Now lets say both the sisters sit on the backseats

NO of options for the front seat= 2 [son or one of the parent]
There has to be one person b/w the two sisters

the arrangements would be like this

No of arrangements= 8

Total no of arrangements = 24+8 = 32

Option B

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by GMATGuruNY » Tue Jan 24, 2012 3:46 am
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by mneeti » Tue Jul 23, 2013 1:11 pm
GuruNY, I have tried a different approach. Not sure, if I got the answer by fluke or the approach is correct. Please help me understand if my logic is correct. Thanks!

Driver seat=2
Rest four seats=4!
Two sisters always sit together = considering 2 sisters as 1 unit = 4c3 *2!

Solution = 2(4!-4c3*2!)=2*16=32
GMATGuruNY wrote:I posted a solution here:

https://www.beatthegmat.com/difficult-pr ... 87921.html

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by [email protected] » Tue Jul 23, 2013 9:04 pm
factor26 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
!
Here's another approach. It's a little longer, but it demonstrates another way to tackle restrictions.

The restriction about the sisters is somewhat problematic, so I decided to ignore the rule and seat all 5 people without obeying that restriction.

Then once I determine the total number of arrangements, I 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 = 32

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com

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by satinder kaur » Thu Jul 25, 2013 5:30 am
Can anybody please help me with this ... I am not able to understand the whole. If I try to apply alphabets order.
Assume Father stands for F , Mother for M , Son for S and Daughters for D.
Now taking the required arrangement of the in first line we can have F/M
in second line any of M/F,D,S,D where D can not come together.
so the possible arrangements for first line : 2!
for second (4! - 4C3 *2) (where two daughters combined to one count)
my concern Is it a right approach?

Thanks

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by ash4gmat » Sun May 01, 2016 11:57 pm
Brent,

How about the way I have done.

1)Seat 1: Either parent seats: 2 ways
2)Seat 3:Any of the 4 seats: 4 ways
3)Seat 4:Any of the pending 3 except sister sits: 2 ways
4)Seat 2:Pending 2 sits: 2 ways
5)Seat 5:Pending 1 sits: 1 way

Answer 32. Correct me if my understanding and method is wrong.

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by [email protected] » Mon May 02, 2016 10:01 am
Hi All,

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