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
The source of this problem is one of the forums I visit and since this relates to a strategy suggested in one of MGMAT's guides, I have posted it here. Please feel free to move it someplace else if this doesnt belong here.
I tried using slots method here.
The driver seat can be filled in 2 ways
The passenger seat can be filled in 4 ways.
The 3 rear seats can be filled in 3*2*1 ways.
All the different ways to fill the seats are : 2*4*3*2*1 = 48
Now, we have to remove all the ways in which two daughters, D1 and D2 sit together. I used the glue method as mentioned in the strategy guide - so, out of 4 possible seats (1 passenger and 3 rear), we have 3 people (considering both daughters as one unit) - of which 3! ways (6) of arranging are possible. Add another 6 to it to consider the possibility of D2 sitting next to D1 - which results in a total of 12 to be removed from the total number of possibilities (48)
48 - 12 = 36.
This is where I am stuck. I think there are a few other combinations to remove from this and I am not sure what they are. Could anyone please let me know if this is the right approach and if so, what the next step is. Any better approaches are definitely welcome.
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- JonathanSchneider
- ManhattanGMAT Staff
- Posts: 370
- Joined: Sun Oct 26, 2008 3:40 pm
The "glue method" doesn't quite apply here, as the four seats are not in a row. Note that your method does not account for the idea that the front passenger seat is not next to any of the back row seats.
If you wish to start with the 48 as you currently have, you must only subtract the cases where the daughters would in fact be next to each other (you have already satisfied the restraint that one of the parents drive). This could only happen when both daughters are in the back seat. The easiest approach is to consider the number of times this would occur, and then find the combinations within each of those occruances.
In short, one of the daughters would have to be in the middle back seat, while the other daughter could be on either side.
This means that there are 2 situations that do not work. For each of those situations, however, the daughters could be in either order, taking us to a total of 4 situations that do not work. But now we must consider that for each of those 4 situations, either of the two parents could be driving, taking us to 8 situations that do not work. And finally, for each of those 8 situations, the remaining parent and the son could switch seats (between the front passenger seat and the remaining vacant rear seat), taking us to a total of 16 situations that do not work. Subtract this 16 from 48, to arrive at the answer = 32.
I think a simpler method, however, is just to start with the slot method, but consider it as follows:
You have two situations that would work: where the daughters are split between the front and back rows; and where the daughters are in the back rows but on opposite sides.
For the first of these two situations, the daughter in the back has 3 options for a seat. You can arrange the daughters in 2 ways. You have 2 choices for which parent drives. And you have 2 choices for the arrangement of the other parent and the son. 3*2*2*2 = 24.
For the second of these situations, you have the same math as above, except that you do not have three options for seating the daughters. They MUST be on either side in the back row. The remaining 2*2*2 still applies, for a total of 8.
Add these two options together for your total: 24 + 8 = 32.
Either way you do it, a hard question.
If you wish to start with the 48 as you currently have, you must only subtract the cases where the daughters would in fact be next to each other (you have already satisfied the restraint that one of the parents drive). This could only happen when both daughters are in the back seat. The easiest approach is to consider the number of times this would occur, and then find the combinations within each of those occruances.
In short, one of the daughters would have to be in the middle back seat, while the other daughter could be on either side.
This means that there are 2 situations that do not work. For each of those situations, however, the daughters could be in either order, taking us to a total of 4 situations that do not work. But now we must consider that for each of those 4 situations, either of the two parents could be driving, taking us to 8 situations that do not work. And finally, for each of those 8 situations, the remaining parent and the son could switch seats (between the front passenger seat and the remaining vacant rear seat), taking us to a total of 16 situations that do not work. Subtract this 16 from 48, to arrive at the answer = 32.
I think a simpler method, however, is just to start with the slot method, but consider it as follows:
You have two situations that would work: where the daughters are split between the front and back rows; and where the daughters are in the back rows but on opposite sides.
For the first of these two situations, the daughter in the back has 3 options for a seat. You can arrange the daughters in 2 ways. You have 2 choices for which parent drives. And you have 2 choices for the arrangement of the other parent and the son. 3*2*2*2 = 24.
For the second of these situations, you have the same math as above, except that you do not have three options for seating the daughters. They MUST be on either side in the back row. The remaining 2*2*2 still applies, for a total of 8.
Add these two options together for your total: 24 + 8 = 32.
Either way you do it, a hard question.
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