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MGMAT Prob
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caramel3536
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Ian Stewart
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First, suppose we had 20 seconds and needed to guess. You can get down to two answers quite quickly. We're making a seating arrangement with 5 people. If there are no restrictions at all, the answer would be 5! = 120. There are restrictions, so the answer must be less. If we ignore the restriction about the daughters, and only note that a parent must be in the driver's seat, we would have 2 choices for that seat, and 4! choices for the rest. That is, we'd have 2*4! = 48 arrangements. But we have the restriction about the daughters, so the answer must be less than 48. That gets us to A) and B) only.
If you want to do this question completely, I don't see any 25-second solution. I think you do need to look at cases. I'll write P, D and S for parent, daughter and son:
1. We could have:
Front: PD
Back: P, D, S
Then we have 2 choices for the driver, and 2 for the other front seat = 4 choices in total (multiply your number of choices). We also have 3*2*1 choices for the back seat, since they can sit in any order. Thus, we have 2*2*3*2*1 = 24 arrangements.
2. We could have:
Front: PP
Back: D, D, S
We have only 2 choices for the front seat (either parent could drive). In the back, the son needs to be in the middle, so there are only 2 arrangements in the back: the daughters could be on either side. There are 2*2 = 4 arrangements.
3. Finally we could have:
Front: PS
Back: D, P, D
Again, we have two choices for the driver. We also have only 2 choices for the back row, because the remaining parent must be in the middle. 2*2 = 4 arrangements.
24+4+4 = 32.
If you want to do this question completely, I don't see any 25-second solution. I think you do need to look at cases. I'll write P, D and S for parent, daughter and son:
1. We could have:
Front: PD
Back: P, D, S
Then we have 2 choices for the driver, and 2 for the other front seat = 4 choices in total (multiply your number of choices). We also have 3*2*1 choices for the back seat, since they can sit in any order. Thus, we have 2*2*3*2*1 = 24 arrangements.
2. We could have:
Front: PP
Back: D, D, S
We have only 2 choices for the front seat (either parent could drive). In the back, the son needs to be in the middle, so there are only 2 arrangements in the back: the daughters could be on either side. There are 2*2 = 4 arrangements.
3. Finally we could have:
Front: PS
Back: D, P, D
Again, we have two choices for the driver. We also have only 2 choices for the back row, because the remaining parent must be in the middle. 2*2 = 4 arrangements.
24+4+4 = 32.
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atlantic
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Ian, great answer. Thanks. I'm new posting here, but I've been lurkling around since one month.
Let me see if I can solve this using straight combinations. It seems to be much more easier.
1. One of the drivers must be either the father or the mother. So let just fix them to the left front seat (sorry, for those living on UK, India, etc, you may like to choose the right front seat); the other seat on the front can be ocupied by on the daughters. Therefore, the three back seats are left for the three remaining members of the family.....
3! x 2 (because on the front seat can be the father or the mother) x 2 (because on the other front seat can be one of the two daughters) = 24
2. Again one of the drivers can be either the father or the mother, but the two daughters must be seated in oposite sides of the back seat. The remaining two members can seat on the remaining seats, one in the front and the middle back seat, resulting in...
2! x 2 (because on the front seat can be the father or the mother) x 2(because the two daughters can change positions) = 8
From 1. and 2. we have 24+8=32. Of course if you draw the seats it will be much more easier to understand. Nevertheless, your 20sec, approach is of really great help. It increases the odds of a smart guessing to 50%.
Let me see if I can solve this using straight combinations. It seems to be much more easier.
1. One of the drivers must be either the father or the mother. So let just fix them to the left front seat (sorry, for those living on UK, India, etc, you may like to choose the right front seat); the other seat on the front can be ocupied by on the daughters. Therefore, the three back seats are left for the three remaining members of the family.....
3! x 2 (because on the front seat can be the father or the mother) x 2 (because on the other front seat can be one of the two daughters) = 24
2. Again one of the drivers can be either the father or the mother, but the two daughters must be seated in oposite sides of the back seat. The remaining two members can seat on the remaining seats, one in the front and the middle back seat, resulting in...
2! x 2 (because on the front seat can be the father or the mother) x 2(because the two daughters can change positions) = 8
From 1. and 2. we have 24+8=32. Of course if you draw the seats it will be much more easier to understand. Nevertheless, your 20sec, approach is of really great help. It increases the odds of a smart guessing to 50%.
