Problem Solving

I've posted this difficult probability question from the MGMAT CAT before, but I still don't get it. Can an expert please analyze my logic and explain why it's wrong?

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?

The first constraint is that one of the parents must drive. So I figured out the possible # of seating arrangements, given this constraint:

2 parents can drive the car
4 people (B, G1,G2, non-driving parent) can sit in the passenger seat
3 people (remaining after filling passenger seat) can sit in left back seat
2 people (remaining after filling left back seat )can sit in middle back seat (persons r
1 person (remaining after filling all other seats) can sit in right back seat

Constraint #1: Total # of seating options: 2*4*3*2*1 = 48

Now, you have to subtract out the second constraint -- the girls can't sit next to each other: (calculated by finding the # of ways the girls DO sit next to each other)

2 parents can drive the car
2 people (Boy, non-driving parent) can sit in the passenger seat [girls must sit in back seat to be next to each other)
3 people (G1, G2, Parent/Boy [whomever wasn't selected for passenger seat]) can sit in left back seat
2 persons (assuming boy was selected for left back seat, either of the girls can sit in the middle)can sit in middle back seat
1 person (remaining girl, after filling all other seats) can sit in right back seat

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

My answer is 48-24= 24

The OA is 32.

What am I doing wrong? Where is my logic flawed? Please specifically point out my error (rather than just showing how to do the problem correctly).

BUMP!!!

Does anyone understand this problem?

Lemme try explaining
there are 3 conditions:
1st condition: a parent driving and a girl sitting in front row:
that leaves with 3 person on rear seats... 3! = 6
now two parents can switch the position that makes it 3! * 2
and two girls can also switch position that makes it 3! * 2 * 2 = 24

2nd Condition: both the girls at the back seat and two parents in front seat so there are 2 ways in which two girls can be seated in teh corner - if dad is driving and mum is sitting besides dad - and also there are 2 ways in which girls can be seated where mum is driving and dad seated besides her so 2*2 ways = 4

3rd condition: two girls at the back seat along with 1 parent
so again 2 ways in which girls can be seated with mum at the back seat and dad driving and again 2 ways in which girls can be seated with dad at the back seat and mum driving so in all 2*2 ways = 4 ways

so 24 + 4+ 4 = 32 ways

Let me know if you still have any question

anju wrote:Lemme try explaining
there are 3 conditions:
1st condition: a parent driving and a girl sitting in front row:
that leaves with 3 person on rear seats... 3! = 6
now two parents can switch the position that makes it 3! * 2
and two girls can also switch position that makes it 3! * 2 * 2 = 24

2nd Condition: both the girls at the back seat and two parents in front seat so there are 2 ways in which two girls can be seated in teh corner - if dad is driving and mum is sitting besides dad - and also there are 2 ways in which girls can be seated where mum is driving and dad seated besides her so 2*2 ways = 4

3rd condition: two girls at the back seat along with 1 parent
so again 2 ways in which girls can be seated with mum at the back seat and dad driving and again 2 ways in which girls can be seated with dad at the back seat and mum driving so in all 2*2 ways = 4 ways

so 24 + 4+ 4 = 32 ways

Let me know if you still have any question

This is much different than the way I did it, because you used factorials. Can you explain what was wrong with my methodology? As you can see, I used the counting method (the # of people available for each individual seat). I believe this method takes the rearrangements into account, eliminating the need for factorials.

If anyone can explain my errors in a simple manner, it would be most helpful. I'm so stressed out over this problem.

Stockmoose16 wrote:Now, you have to subtract out the second constraint -- the girls can't sit next to each other: (calculated by finding the # of ways the girls DO sit next to each other)

2 parents can drive the car
2 people (Boy, non-driving parent) can sit in the passenger seat [girls must sit in back seat to be next to each other)
3 people (G1, G2, Parent/Boy [whomever wasn't selected for passenger seat]) can sit in left back seat
2 persons (assuming boy was selected for left back seat, either of the girls can sit in the middle)can sit in middle back seat
1 person (remaining girl, after filling all other seats) can sit in right back seat

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

working on the second part

2 parents can drive the car
2 people (Boy, non-driving parent) can sit in the passenger seat [girls must sit in back seat to be next to each other)
2 people (G1, G2) can sit in middle back seat. (If girls dont sit in the middle seat then they wont be together)
2 persons (one of the remaining girls, and boy/parent) can sit in the side seat
1 person (remaining person) can sit in right back seat

=2*2*2*2*1= 16

fault in your assumption
3 people (G1, G2, Parent/Boy [whomever wasn't selected for passenger seat]) can sit in left back seat
2 persons (assuming boy was selected for left back seat, either of the girls can sit in the middle)can sit in middle back seat

you took any 3 for left back seat, and then assumed that boy was sitting in left back seat............contradiction

Ian Stewart wrote:

Stockmoose16 wrote:I've posted this difficult probability question from the MGMAT CAT before, but I still don't get it. Can an expert please analyze my logic and explain why it's wrong?

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?

The first constraint is that one of the parents must drive. So I figured out the possible # of seating arrangements, given this constraint:

2 parents can drive the car
4 people (B, G1,G2, non-driving parent) can sit in the passenger seat
3 people (remaining after filling passenger seat) can sit in left back seat
2 people (remaining after filling left back seat )can sit in middle back seat (persons r
1 person (remaining after filling all other seats) can sit in right back seat

Constraint #1: Total # of seating options: 2*4*3*2*1 = 48

Now, you have to subtract out the second constraint -- the girls can't sit next to each other: (calculated by finding the # of ways the girls DO sit next to each other)

2 parents can drive the car
2 people (Boy, non-driving parent) can sit in the passenger seat [girls must sit in back seat to be next to each other)
3 people (G1, G2, Parent/Boy [whomever wasn't selected for passenger seat]) can sit in left back seat
2 persons (assuming boy was selected for left back seat, either of the girls can sit in the middle)can sit in middle back seat
1 person (remaining girl, after filling all other seats) can sit in right back seat

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

My answer is 48-24= 24

The OA is 32.

What am I doing wrong? Where is my logic flawed? Please specifically point out my error (rather than just showing how to do the problem correctly).

This is a more labour-intensive counting problem than I've ever seen on a real GMAT, so don't stress about it too much. You almost got to the correct answer, with one small error near the end: when you count the number of ways for both girls to be seated together, you haven't actually made sure that both girls sit together. That is, when you do the following:

2 parents can drive the car
2 people (Boy, non-driving parent) can sit in the passenger seat [girls must sit in back seat to be next to each other)
3 people (G1, G2, Parent/Boy [whomever wasn't selected for passenger seat]) can sit in left back seat
2 persons (assuming boy was selected for left back seat, either of the girls can sit in the middle)can sit in middle back seat
1 person (remaining girl, after filling all other seats) can sit in right back seat

some of the arrangements you are counting will have one girl in the left back seat, and the other girl in the right back seat. You could adjust as follows:

-if the two girls *will* sit together, one girl *must* be in the middle back seat (in the above, you've allowed the boy/parent to sit in the middle back seat). So we have:

* 2 parents can drive
* 2 people for the front passenger seat (boy or other parent)
* 2 girls could sit in the middle back seat
* 2 choices for the left back seat (the other girl, or the boy/parent)
* 1 choice remains for the right back seat

and there are therefore 2*2*2*2 = 16 ways to seat the two girls together. Now, as you did, you need to subtract: 48-16 = 32 seating arrangements in total which satisfy the conditions given in the question.

Ian,

I can't thank you enough for this explanation. Your posts are always so clear and helpful. Thank you for explaining why my methodology is incorrect. Would you mind also answering the other P/C I posted, explaining why my methodology is right/wrong. Here's the link:
= https://www.beatthegmat.com/tough-gmat ... t9484.html

is this a correct way to think?

there's drivers seat for 1 of the 2parents. rest there are 4 seats.
number of ways to sit in 4 seats is 4! & then there are 2 parents who can be drivers so total way to seat is 2*4! = 48

now way in which 2 girls can sit :
they can only sit in back set there are 4 ways to do this in 3 seats.
g1g2x , g2g1x , xg1g2 , xg2g1 x -can be boy or one parent
the remaining one back seat can be occupied by boy or parents - 2 ways
& then 2 ways for driver's seat

that gives 2*2*4 =16

so 48-16 =32

Split into two parts a daughter sits in the front passenger seat
a daughter does not sit in the front.

D1, D2 - > two daughters
M - Mother
F - Father
S - son
{} - a seat
A) One Daughter sits in the front

Front row -> {M,F} {D1,D2} - 2*2
Back row -> 3 people remain - 3! arrangements
Total -> 2*2*3! = 24

B) Both daughters in the back

Front row-> {M,F} {M or F, S} = 2*2

Back row -> {D1,D2} {M or F or S} {D1 or D2} = 2*1*1 = 2

Total -> 2*2*2 = 8

Grand total 8+ 24 = 32