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5 people run a race and

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by netcaesar » Mon Sep 07, 2009 1:38 pm
5 people run a race and none of them arrives at the same time. In how many ways can the race end with Bob ahead of Marge?

OA is 10
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by tom4lax » Mon Sep 07, 2009 2:00 pm
I would actually think that it's 60.

Total ways the race can finish is 5! = 120. Now divide by two because whatever order of Bob and Marie, only one of them is valid.

How did you arrive at 10?
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by netcaesar » Mon Sep 07, 2009 2:42 pm
5!/2!3!
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by tom4lax » Mon Sep 07, 2009 2:51 pm
Whats your logic behind 3!2!?
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by Nermal » Tue Sep 08, 2009 12:04 am
I also calculated it the way tom4lax did.

What came to my mind afterwards was:
Does "Bob ahead of Marge" mean that he finishes directly before Marge does or could it also be that Bob finishes first and Marge last?
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by netcaesar » Tue Sep 08, 2009 6:54 am
Sorry, you are right.
If there are 120 combinations, in half of them Bob is ahead of Marge, so the answer is 60.
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by txeconomist » Thu Sep 10, 2009 2:00 pm
If it doesnt matter where people other than Bob and Marge place, then its 10 I guess.

BMMMM
XBMMM
XXBMM
XXXBM
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by Mayur Sand » Thu Sep 10, 2009 7:17 pm
My solution is following,
Lets suppose following are finished positions
p1,p2,p3,p4 ,p5 . Now If Bob came at p1 then Marley can come at p2,p3,p4,p5 which means 4 ways , similary when Bob comes at p2 then Marley can come at p3,p4,p5 3 ways
so it should be 4*3*2*1 = 24 ways

Let me know my mistake here
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by prindaroy » Sat Sep 12, 2009 3:07 pm
The answer should be 60. The answer 10 is very silly. The total number of possibilities = 120 and in half of them, Bob will finish ahead of Marge. So the answer is 60. 10 cannot be the answer because there are 5 different people in 5 different positions.
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