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# JOhn and Bill are among five runners in a race where there

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Manager
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JOhn and Bill are among five runners in a race where there [#permalink]

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17 Apr 2008, 12:20
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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
Intern
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17 Apr 2008, 12:32
i think the answer shud be 60 .
By the way wot is the answer
CEO
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Concentration: Entrepreneurship, Other
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17 Apr 2008, 12:38
D

1. All possible combinations: $$P^5_5=5!=120$$

2. Due to symmetry, all combinations can be evenly divided by two groups: John ahead of Bill and Bill ahead of John. So, $$N=\frac{120}{2}=60$$
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Manager
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17 Apr 2008, 15:11
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..
Manager
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17 Apr 2008, 16:03
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?
SVP
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17 Apr 2008, 18:28
im getting 10 as well, by just listing out the possibilities ...
Director
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17 Apr 2008, 23:32
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
Director
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18 Apr 2008, 12:40
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
VP
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18 Apr 2008, 13:15
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
Re: race   [#permalink] 18 Apr 2008, 13:15
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