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by maihuna » Sun Aug 09, 2009 3:53 am
A railway carriage will accommodate 5 passengers on each side. In how many ways can ten persons take their seats when two of them can not face the engine and a third can not travel backwards.

720
1440
2400
2520
504000
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by DanaJ » Sun Aug 09, 2009 4:19 am
I'm not very good at this, so take my explanation with a grain of salt. I'd also love to see other opinions!
Out of the ten spots, you get:
- three "pretentious" passengers
- seven mobile passengers

Let's tackle the first three:
a. the guy who can't travel backwards is fixed on the front-facing side, giving him 5 possible positions on that one
b. the two who won't travel facing the engine will have to sit on the other side. Since they're picky, they'll probably want to be seated first and will choose between the 5 chairs available: 5P2 = 20

The other seven passengers will sit in the remaining seven seats. They can do so in 7! = 5040 ways.

So I got 5*20*5040 = 504000 ways. Of course, I realize that this is an insanely large number, so, again, I may be wrong.
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by maihuna » Sun Aug 09, 2009 10:18 am
Its correct DanaJ.

What about out of these 10, two A and B has to sit facing each other.
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by DanaJ » Sun Aug 09, 2009 11:49 am
Well, if that's the only restriction, then it's not that hard, I suppose.

A and B make up a pair and the pair will sit in one of 5 positions. Since you can switch between A and B, then there are 2*5 = 10 different ways to arrange them in the seats. The remaining passengers will be seated in 8! ways, i.e. 40320.

The end result will be 10*40320 = 403200.

As was the case with 7!, I calculated 8! with the Win Calculator. IMHO, such laborious calculations are unlikely on the GMAT: at an advanced level, it's more about wit/tricks/attention then sheer labor!
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by maihuna » Sun Aug 09, 2009 12:01 pm
The answer, as is the case here, could be in following format: !10/9

The same as : 10*!8 = 10*9*!8/9 = !10/9
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