BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Meg and Bob

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Junior | Next Rank: 30 Posts
Posts: 15
Joined: Thu Jul 02, 2009 9:53 pm

Meg and Bob

by beatthegmat2910 » Mon Nov 30, 2009 11:21 pm
Meg and Bob are among the 5 participants in a cycling race. If each participant finishes the race and no two participants finish at the same time, in how many different possible orders can the participants finish the race so that Meg finishes ahead of Bob?
A. 24
B. 30
C. 60
D. 90
E. 120


Please help.
Top
Source: — Problem Solving |
User avatar
Community Manager
Posts: 1537
Joined: Mon Aug 10, 2009 6:10 pm
Thanked: 653 times
Followed by:252 members

by papgust » Tue Dec 01, 2009 12:53 am
Interesting question!

Totally 5 players. Meg, Bob and other 3 players.
Meg has to be ahead of Bob in whatever position Meg is. Consider the following scenarios,

(1) M _ _ _ _ - Meg is in first place. Bob and other players can occupy in other 4 places.
So 4! = 24

OR

(2) _ M _ _ _ - Meg is in second place. 1 of other 3 players must occupy first place. Bob and other 2 players can occupy in any 3 positions following Meg.
So, 3 * 3! = 18

OR

(3) _ _ M _ _ - Meg is in third place. 2 of other 3 players must occupy 1st and 2nd places. Bob and other 1 player can occupy in any 2 positions following Meg.
So, 3 * 2 * 2! = 12

OR

(4) _ _ _ M _ - Meg is in fourth place. Other 3 players must occupy first 3 places. Bob can occupy only the last place.
So, 3 * 2 * 1 * 1 = 6

Meg CANNOT be in fifth place because Meg has to ahead of Bob at any time.

Add all 4, 24+18+12+6 = 60.

Choose C.
Top
GMAT Instructor
Posts: 1302
Joined: Mon Oct 19, 2009 2:13 pm
Location: Toronto
Thanked: 539 times
Followed by:164 members
GMAT Score:800

by Testluv » Tue Dec 01, 2009 8:27 am
How many ways are there of arranging them without the restriction that Meg finishes ahead of Bob?

Well, the number of ways of arranging n objects is n!

So, there are 5! = 120 ways of arranging these five racers without considering the restriction.

Now, let's consider the restriction. In how many of these 120 ways will Meg finish ahead of Bob? In exactly half of the ways. For example, if Meg finished ahead of Bob in more than half of the total possible orders, then that would would mean Bob finishes ahead of Meg less than half the time, and that is absurd. In the total number of orders Meg finishes ahead of Bob half the time, and Bob finishes ahead of Meg the other half of the times.

Therefore, the final answer is just 120/2 = 60.
Kaplan Teacher in Toronto
Top
Junior | Next Rank: 30 Posts
Posts: 15
Joined: Thu Jul 02, 2009 9:53 pm

by beatthegmat2910 » Tue Dec 01, 2009 10:48 am
Hey thanks Testluv..this solution is much easier and requires minimal calculation.
Top
Post new topic Post Reply
• Page 1 of 1