5 people run a race and

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netcaesar: Junior | Next Rank: 30 Posts; Posts: 29; Joined: Thu Sep 25, 2008 5:11 pm

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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Source: Beat The GMAT — Problem Solving | Mon Sep 07, 2009 1:38 pm

tom4lax: Master | Next Rank: 500 Posts; Posts: 175; Joined: Mon Feb 09, 2009 3:57 pm; Thanked: 4 times

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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netcaesar: Junior | Next Rank: 30 Posts; Posts: 29; Joined: Thu Sep 25, 2008 5:11 pm

by netcaesar » Mon Sep 07, 2009 2:42 pm

5!/2!3!

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tom4lax: Master | Next Rank: 500 Posts; Posts: 175; Joined: Mon Feb 09, 2009 3:57 pm; Thanked: 4 times

by tom4lax » Mon Sep 07, 2009 2:51 pm

Whats your logic behind 3!2!?

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Nermal: Senior | Next Rank: 100 Posts; Posts: 77; Joined: Sun Jun 21, 2009 10:25 am; Location: Germany; Thanked: 7 times

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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netcaesar: Junior | Next Rank: 30 Posts; Posts: 29; Joined: Thu Sep 25, 2008 5:11 pm

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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txeconomist: Junior | Next Rank: 30 Posts; Posts: 12; Joined: Mon Jul 07, 2008 7:28 am

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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Mayur Sand: Master | Next Rank: 500 Posts; Posts: 166; Joined: Fri Apr 03, 2009 11:16 pm; Location: India

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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prindaroy: Master | Next Rank: 500 Posts; Posts: 113; Joined: Thu Jul 16, 2009 11:23 am; Thanked: 15 times; GMAT Score:730

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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