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Re: JOhn and Bill are among five runners in a race where there [#permalink]
puma wrote:
how many ways is it possible for John to finish the race ahead of Bill?


(B) for me. If I get the question right, we care only about John's and Bill's positions. Then we have:
J B 3 4 5
J 2 B 4 5
J 2 3 B 5
J 2 3 4 B

1 J B 4 5
1 J 3 B 5
....
i.e. 4+3+2+1 = 10 possible ways..
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Re: JOhn and Bill are among five runners in a race where there [#permalink]
For me:

J 4 3 2 1= 4!= 24

3 J 3 2 1= 3*3!= 9

which means it is greater than 30.

And 5! = 120, therefore it has to be less than 120.

Only 60 fits the bill.

Whats the OA/OE?
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Re: JOhn and Bill are among five runners in a race where there [#permalink]
im getting 10 as well, by just listing out the possibilities ...
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Re: JOhn and Bill are among five runners in a race where there [#permalink]
If we care about the other participants' positions then the answer is 60 pointed out by walker,

if we dont (which is the case here) we have to divide this be 3! (the number of ways the others can finish).

so it is infact 60/6 = 10
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Re: JOhn and Bill are among five runners in a race where there [#permalink]
puma wrote:
OA is D


this is so cruel :-)

hmm...i am a bit confused either way. On the one hand I am not sure if "how many ways is it possible for John to finish the race ahead of Bill" implies that the other runner's positions matter
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Re: JOhn and Bill are among five runners in a race where there [#permalink]
puma wrote:
JOhn and Bill are among five runners in a race where there are no ties. how many ways is it possible for John to finish the race ahead of Bill?

a) 5
b) 10
c) 30
d) 60
e) 120


D.

Layout the options
JBXXX = 4! = 24
XJBXX = 3*3! = 18
XXJBX = 2*3! = 12
XXXJB = 3! = 6

Total = 24+18+12+6 = 60



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