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Pat will walk from intersection X to intersection Y along a

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Pat will walk from intersection X to intersection Y along a [#permalink] New post 01 Aug 2003, 07:36
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Pat will walk from intersection X to intersection Y along a route that is confined to the square grid of four streets and three avenues shown in the map above. How many routes from X to Y can Pat take that have the minimum possible length?



Can you show me the best way to attack path questions?
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 [#permalink] New post 01 Aug 2003, 11:19
Three Ups ( U) and two side roads (S)
So possible routes with minimum possible length = 5! / (3! * 2!) = 10
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Hi Stolyar [#permalink] New post 01 Aug 2003, 11:54
Will come back to that other post on "GMAT Classic" later in the show.

Got something stuck in my head with expressing this simple division conretely. :-D

Okay, Kpadama,

120 ways to travel this route
-------------------------------
12 ways to travel the shortest.

So every 12th way there is a shortest route. This brings the total to 10 different minimum routes.
this sounds so foreign 8-)

Stolyar, I think your way is just a coincidence. Twist around the number of up and down roads and is quite a different way.

8-)

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Re: Hi Stolyar [#permalink] New post 01 Aug 2003, 11:56
Curly05 wrote:
Will come back to that other post on "GMAT Classic" later in the show.

Got something stuck in my head with expressing this simple division conretely. :-D

Okay, Kpadama,

120 ways to travel this route
-------------------------------
12 ways to travel the shortest.

So every 12th way there is a shortest route. This brings the total to 10 different minimum routes.
this sounds so foreign 8-)

Stolyar, I think your way is just a coincidence. Twist around the number of up and down roads and is quite a different way.

8-)



Stolyar is exactly correct.

If you do not take the shortest route than there are an infinite number of ways to get from A to B.
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Hi [#permalink] New post 01 Aug 2003, 12:28
Okay, Kpadama,

120 ways to travel this route
-------------------------------
12 ways to travel the shortest.

So every 12th way there is a shortest route. This brings the total to 10 different minimum routes.
this sounds so foreign

This sounds funny, Akami, can you clear it up Mr.Kpadama way.
VT :wink:
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My problem with it [#permalink] New post 01 Aug 2003, 12:44
There are 3 roads from Pittsburgh to Ocean City, Md and four roads from Ocean city, Md. to Bay City,CA. If Allison drives from Pittsburgh to Bay City, CA and back, passes from Ocean City in both directions, and does not travel any road twice, how many different routes for the trip are possible?
You see in this similar problem, can anybody come up with permus or combos to solve it.

So, getting back to Kpadama's way, I guess we state there are five diff. roads. We start with the first road and can go five different ways, :wink: . I'm lost here.

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Re: Hi Stolyar [#permalink] New post 01 Aug 2003, 13:36
AkamaiBrah wrote:
Curly05 wrote:
Will come back to that other post on "GMAT Classic" later in the show.

Got something stuck in my head with expressing this simple division conretely. :-D

Okay, Kpadama,

120 ways to travel this route
-------------------------------
12 ways to travel the shortest.

So every 12th way there is a shortest route. This brings the total to 10 different minimum routes.
this sounds so foreign 8-)

Stolyar, I think your way is just a coincidence. Twist around the number of up and down roads and is quite a different way.

8-)



Stolyar is exactly correct.

If you do not take the shortest route than there are an infinite number of ways to get from A to B.



Dear AkamaiBrah & stolyar,

I beg to differ, even though I consider you both my Guru's, that if the
number are changed from 3&2 to 7&2, your approach will fail. In addition,
I don't think we can treat this problem as a simple combination one.
I would like to get your feedback?
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 [#permalink] New post 01 Aug 2003, 13:42
Dear AkamaiBrah & stolyar,

After posting the previous response, I realised that I was wrong. I think,
I am forgetting even simple addition and basic logic. :oops:
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 [#permalink] New post 01 Aug 2003, 13:48
kpadma wrote:
Dear AkamaiBrah & stolyar,

After posting the previous response, I realised that I was wrong. I think,
I am forgetting even simple addition and basic logic. :oops:


That's okay. By daring to challenge us, then realizing after thinking more carefully that you are wrong, you undoubtably understand this problem (and combinations in general) much better than if you were to just accept what we say and move on. It is good that you took the time to turn this around in your head a few times.

Good job.
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Coming to some sort of consensus [#permalink] New post 01 Aug 2003, 13:59
So whose way is right?

Kpadama,

Explain your way. We need to come to some sort of conclusion or consensus with this problem.

Why do you say Stoylar's way is wrong?

Akami- Dude, you are the genius who can resolve the arguing I started.

VT
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Hi [#permalink] New post 02 Aug 2003, 06:52
Son, Akami, can you resolve the argument here.

Yes, can't you see the problem with Stoylar's approach?

Why don't we multiply combinations here, this raises a good intellectual discussion :wink:
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request explanation [#permalink] New post 18 Aug 2003, 12:29
kpadma wrote:
Three Ups ( U) and two side roads (S)
So possible routes with minimum possible length = 5! / (3! * 2!) = 10


Request explanation in detail please.

thanks

praetorian
request explanation   [#permalink] 18 Aug 2003, 12:29
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