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# pls provide the detailed solution

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VP
Joined: 18 May 2008
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pls provide the detailed solution [#permalink]  23 Jun 2008, 22:25
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0% (00:00) correct 0% (00:00) wrong based on 0 sessions
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pls provide the detailed solution
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shortest path.doc [25.5 KiB]

SVP
Joined: 17 Jun 2008
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Re: PS: shortest path [#permalink]  23 Jun 2008, 22:48

If I name the various corners as follows

10 11 12
7 8 9
4 5 6
1 2 3

Then, the following are the possible routes (the shortest routes will be going horizontal and vertical only):

1-2-3-6-9-12
1-2-5-6-9-12
1-2-5-8-9-12
1-2-5-8-11-12
1-4-5-6-9-12
1-4-5-8-9-12
1-4-5-8-11-12
1-4-7-8-9-12
1-4-7-8-11-12
1-4-7-10-11-12

If there is any short cut to this, please let me know.
Current Student
Joined: 12 Jun 2008
Posts: 287
Schools: INSEAD Class of July '10
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Re: PS: shortest path [#permalink]  24 Jun 2008, 00:48

To go from X to Y, you have to take 5 segments. 3 vertical segments and 2 horizontal segments in any order you want.

So this is just a matter of choosing (for instance) which of the 2 segments amongst the 5 will be horizontal segments.

Number is $$2C5=\frac{5!}{3!*2!}=10$$
VP
Joined: 18 May 2008
Posts: 1287
Followers: 14

Kudos [?]: 240 [0], given: 0

Re: PS: shortest path [#permalink]  24 Jun 2008, 02:14
Wow Oski ! u rock! Answer is 10 indeed. thanks a lot!
Oski wrote:

To go from X to Y, you have to take 5 segments. 3 vertical segments and 2 horizontal segments in any order you want.

So this is just a matter of choosing (for instance) which of the 2 segments amongst the 5 will be horizontal segments.

Number is $$2C5=\frac{5!}{3!*2!}=10$$
Re: PS: shortest path   [#permalink] 24 Jun 2008, 02:14
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# pls provide the detailed solution

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