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# John and Mike are among the runners of a 5 people foot-race.

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VP
Joined: 06 Feb 2007
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John and Mike are among the runners of a 5 people foot-race. [#permalink]  01 Apr 2007, 14:23
John and Mike are among the runners of a 5 people foot-race. How many possible finishes are there if John must arrive in front of Mike?
A) 3
B) 6
C) 18
D) 60
E) 120
Manager
Joined: 25 May 2006
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Getting D. Got to go, but if its correct will explain later.
_________________

Who is John Galt?

Manager
Joined: 26 Feb 2007
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total no. of possible ways of finishing the race = 5 P5 =120, Out of this, probability of one person finishing ahead of other is 1/2 so there are 120*1/2 =60 ways

hence D
Manager
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Re: Runners [#permalink]  01 Apr 2007, 17:08
nervousgmat wrote:
John and Mike are among the runners of a 5 people foot-race. How many possible finishes are there if John must arrive in front of Mike?
A) 3
B) 6
C) 18
D) 60
E) 120

Simple way:
There are 120 ways of finishing the race. Out of this 1/2 the time A will be ahead of B hence answer is 60.

Another solution:
since there are 5 people, you can say that remaining 3 people can be arranged in 3! = 6 ways. Now if A comes first then B can come in 4 different positions. If A comes second B can come in 3 different positions. Continuing this way you will see that there are 4+3+2+1 = 10 ways in which B is after A. Multiply this by 6 to get 60
VP
Joined: 25 Jun 2006
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for Mike and John, there are only 2 possibilities:
MIke in front of John
Mike behind John.

So the question is asking for 1 of these 2 cases. So it is permutation divided by 2.

Senior Manager
Joined: 11 Feb 2007
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Yes D.

I drew five circles each representing a runner...

Didn't take me long, but primitive compared to how others did.

If J is first, then there are 24 ways with M behind J.

If J is second, then there are 18 ways with M behind J.

If J is third (middle) then 12 ways.

J fourth (hence M last) then 6 ways.

24+18+12+6 = 60

VP
Joined: 06 Feb 2007
Posts: 1023
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