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# Pat will walk from intersection A to intersection B along a

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Pat will walk from intersection A to intersection B along a [#permalink]  03 Aug 2008, 19:22
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Pat will walk from intersection A to intersection B 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 A to B can Pat take that have the minimum possible length?

A. 6
B. 8
C. 10
D. 14
E. 16

[Reveal] Spoiler: OA

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File comment: map
PS combinatorics.doc [26.5 KiB]

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Re: PS: Combinatorics [#permalink]  04 Aug 2008, 05:02
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haidzz wrote:
Pat will walk from intersection A to intersection B 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 A to B can Pat take that have the minimum possible length?
A. 6
B. 8
C. 10
D. 14
E. 16

(map is attached)

Attachment:
PS combinatorics.doc

This is a question from the OG-11. Does anyone have a faster/easier solution??

Consider this a as coordinate plane with A as origin and B as (2,3).

The shortest possible way is when you go take either right or up. (lefts and downs make repetitions and thus non shortest paths)

now, to get to (2,3) form (0,0) you need min 5 steps.

so you have to choose all the paths such that you have 2 steps out of 5 along x axis and getting to 3 of y axis.
The number of ways = 5C2 =10

(or alternately all paths that have 3 steps along y and takes you to 2 of x. .. 5C3 = 10)
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Re: PS: Combinatorics [#permalink]  04 Aug 2008, 06:25
It is a fairly simple problem if one recognizes that there's only one combination of 3 y-coordinates for every pair of x-coordinates. In simpler terms, you can go right in 2 paths, but there's only one unique set of 3 paths going upwards with those 2 paths.
All we have to do is figure out in how many ways we can take 2 paths to the right. For each such path, there will be only 1 combination of 3 paths that will go up to B.
Also note that there are 8 paths going to the right but we cannot do 8C2 because once you take the 1st path going to the right, all the other right-paths above it become redundant.
The number of paths we can take to the right is 1*4 + 1*3 + 1*2 + 1*1 = 10 paths.
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Re: PS: Combinatorics [#permalink]  04 Aug 2008, 06:43
haidzz wrote:
dude you're a genius! kudos to you. I'm really weak in Permutations, combinations and probability. How can I get better at it? Can you suggest a particular book other than GMAT books?

This site is pretty good...
http://www.mansw.nsw.edu.au/members/ref ... no4yen.htm

Thanks to Durgesh79 for pointing me to this page. Awesome summary of all basic concepts.

Also follow his solutions to combinatorial problems.. he comes up with real cool ways to tame these little pups.
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Re: PS: Combinatorics [#permalink]  08 Jul 2010, 18:55
Bringing up an old topic because it is giving me headaches.

I understand breaking this down into 5 steps (2 Rights, 3 Ups)

So right = R
And Up = U

So we need to see how many permutations include 2 Rights, and 3 Ups

Start to the right
RRUUU
RURUU
RUURU
RUUUR

URRUU
URURU
URUUR
UURRU
UURUR
UUURR

Obviously making a list is not the quickest way to solve this problem.

How do I get to it this ----> 5!/2!3! = 10 ????????

It seems to be me are picking 5 out of 5? So we would have 5!/5!0! which equals 1
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Re: PS: Combinatorics [#permalink]  08 Jul 2010, 20:55
16
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Quote:
How do I get to it this ----> 5!/2!3! = 10 ????????

by using the formula for arranging n objects where some objects recur. We know that there are n! ways of arranging n distinct objects. So, for example, the number of words (both sensical and nonsensical) that can be made from the letters in this word:

GMAT

is 4!.

What about the number of words that can be created from this word:

DESERT?

Well, we have 6 objects. If they were all distinct, there would be 6! ways of arranging all the letters, and thus 6! words would be made. HOWEVER, not all of the letters are distinct. In particular, "E" shows up twice. So, there are actually 6!/2! words we can create. What about this word:

DESSERT?

Now, we have 7 objects. But both "S" and "E" show up twice. So, there are 7!/(2!*2!) ways of arranging.

DESSERTS?

Now, there are 8!/(3!*2!) ways of arranging.

SSS

Well, there are 3!/3! or 1 way of arranging all the letters.

The formula for arranging n objects where some objects recur is: n!/(r!*s!) in which "r" and "s" are the number of times objects of a certain kind appear.

So, with:

UUURR,

there are 5!/(3!*2!) ways of arranging.
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Prepare with Kaplan and save $150 on a course! Kaplan Reviews Manager Status: Waiting to hear from University of Texas at Austin Joined: 24 May 2010 Posts: 76 Location: Changchun, China Schools: University of Texas at Austin, Michigan State Followers: 5 Kudos [?]: 41 [0], given: 4 Re: PS: Combinatorics [#permalink] 08 Jul 2010, 21:10 Quote: The formula for arranging n objects where some objects recur is: n!/(r!*s!) in which "r" and "s" are the number of times objects of a certain kind appear. Thanks for your help, just to make sure I understand if I had MISSISSIPPI The number of ways to arrange it would be 11! / (4!*4!*2!) = 34,650 ??? Kaplan GMAT Instructor Joined: 21 Jun 2010 Posts: 75 Location: Toronto Followers: 23 Kudos [?]: 117 [1] , given: 2 Re: PS: Combinatorics [#permalink] 08 Jul 2010, 21:12 1 This post received KUDOS TallJTinChina wrote: Quote: The formula for arranging n objects where some objects recur is: n!/(r!*s!) in which "r" and "s" are the number of times objects of a certain kind appear. Thanks for your help, just to make sure I understand if I had MISSISSIPPI The number of ways to arrange it would be 11! / (4!*4!*2!) = 34,650 ??? Haha! That's probably the best example one can come up with. Yes, that is indeed correct....thanks for making me count all the "I"s and "S"s! _________________ Kaplan Teacher in Toronto http://www.kaptest.com/GMAT Prepare with Kaplan and save$150 on a course!

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Re: PS: Combinatorics [#permalink]  21 Jan 2011, 16:42
bhushangiri wrote:
haidzz wrote:
Pat will walk from intersection A to intersection B 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 A to B can Pat take that have the minimum possible length?
A. 6
B. 8
C. 10
D. 14
E. 16

(map is attached)

Attachment:
PS combinatorics.doc

This is a question from the OG-11. Does anyone have a faster/easier solution??

Consider this a as coordinate plane with A as origin and B as (2,3).

The shortest possible way is when you go take either right or up. (lefts and downs make repetitions and thus non shortest paths)

now, to get to (2,3) form (0,0) you need min 5 steps.

so you have to choose all the paths such that you have 2 steps out of 5 along x axis and getting to 3 of y axis.
The number of ways = 5C2 =10

(or alternately all paths that have 3 steps along y and takes you to 2 of x. .. 5C3 = 10)

Can you explicate your combinatorial method a bit more. For example, If I add another Avenue, eg, Avenue D, and another street, Street 5, I will have to go 3 steps right and 4 steps up to get to point C.

Therefore, the total number of route will be 7C3 = 35? Am I correct?
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Re: PS: Combinatorics [#permalink]  21 Jan 2011, 16:56
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Expert's post
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ezinis wrote:
bhushangiri wrote:
haidzz wrote:
Pat will walk from intersection A to intersection B 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 A to B can Pat take that have the minimum possible length?
A. 6
B. 8
C. 10
D. 14
E. 16

(map is attached)

Attachment:
The attachment PS combinatorics.doc is no longer available

This is a question from the OG-11. Does anyone have a faster/easier solution??

Consider this a as coordinate plane with A as origin and B as (2,3).

The shortest possible way is when you go take either right or up. (lefts and downs make repetitions and thus non shortest paths)

now, to get to (2,3) form (0,0) you need min 5 steps.

so you have to choose all the paths such that you have 2 steps out of 5 along x axis and getting to 3 of y axis.
The number of ways = 5C2 =10

(or alternately all paths that have 3 steps along y and takes you to 2 of x. .. 5C3 = 10)

Can you explicate your combinatorial method a bit more. For example, If I add another Avenue, eg, Avenue D, and another street, Street 5, I will have to go 3 steps right and 4 steps up to get to point C.

Therefore, the total number of route will be 7C3 = 35? Am I correct?
Attachment:

Q.png [ 25.04 KiB | Viewed 17364 times ]

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?
A) 6
B) 8
C) 10
D) 14
E) 16

In order the length to be minimum Pat should only go UP and RIGHT: namely thrice UP and twice RIGHT.

So combination of UUURR: # of permutations of 5 letters out of which there are 3 identical U's and 2 identical R's is 5!/3!2!=10.

If there were 5 streets and 4 avenues then the answer would be combination of UUUURRR: # of permutations of 7 letters out of which there are 4 identical U's and 3 identical R's is 7!/4!3!=35.

Similar questions:
grockit-similar-to-og-quant-qustion-99962.html
casey-and-the-bus-104236.html
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Re: PS: Combinatorics [#permalink]  21 Jan 2011, 17:19
You're the best Bunuel, +1
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Re: Pat will walk from intersection A to intersection B along a [#permalink]  13 Apr 2012, 10:11
I'm not able to understand this one. Could someone explain? thanks.
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Re: Pat will walk from intersection A to intersection B along a [#permalink]  13 Apr 2012, 10:18
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I'm not able to understand this one. Could someone explain? thanks.

There are several solutions above, so can you please specify a bit what didn't you understand?
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Re: Pat will walk from intersection A to intersection B along a [#permalink]  13 Apr 2012, 10:31
I didn't understand this part "In order the length to be minimum Pat should only go UP and RIGHT: namely thrice UP and twice RIGHT.". He could also go twice RIGHT and then thrice UP. I didn't follow how this was chosen.
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Re: Pat will walk from intersection A to intersection B along a [#permalink]  13 Apr 2012, 10:44
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I didn't understand this part "In order the length to be minimum Pat should only go UP and RIGHT: namely thrice UP and twice RIGHT.". He could also go twice RIGHT and then thrice UP. I didn't follow how this was chosen.

Y is 3 moves UP and 2 moves RIGHT from X (Pat). Now, in order to minimize the route from X to Y only those moves should be taken, how else? But Pat can make these moves (3 Ups, 2 RIGHT) in several different ways:
Up, Up, Up, Right, Right;
Up, Up, Right, Up, Right;
Right, Up, Up, Right, Up;
...
Just look at the diagram to check.

Now, how, many combinations of those moves are possible? # of combination of Up, Up, Up, Right, Right or combinations UUUPP is 5!/(3!2!)=10.

Hope it's clear.
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Re: Pat will walk from intersection A to intersection B along a [#permalink]  29 Jul 2012, 15:35
Bunuel, great example & explanation! Thank you
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Re: Pat will walk from intersection A to intersection B along a [#permalink]  30 Jul 2012, 02:24
1
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haidzz wrote:
Pat will walk from intersection A to intersection B 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 A to B can Pat take that have the minimum possible length?
A. 6
B. 8
C. 10
D. 14
E. 16

(map is attached)

Attachment:
PS combinatorics.doc

This is a question from the OG-11. Does anyone have a faster/easier solution??

Looking at the map, we can see that a route of minimal length is any route that takes only steps up (U) and to the right(R), never to the left or down.
Necessarily, such a route has two Right walks and three Up walks. One have to count all the orderings to have in a sequence of 5 walks 2Rs and 3Us.
So, there are 5C2 possibilities for a minimal route, which is 5*4/2=10.

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Re: Pat will walk from intersection A to intersection B along a [#permalink]  21 Jan 2013, 01:47
Great explanation Bunuel, thanks a lot.
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Re: Pat will walk from intersection A to intersection B along a [#permalink]  15 Sep 2013, 11:18
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2V3H

Total - 5

so, 5!/3!2! = 10

Ans:c
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Re: PS: Combinatorics [#permalink]  23 Apr 2014, 18:25
Testluv wrote:
Quote:
How do I get to it this ----> 5!/2!3! = 10 ????????

by using the formula for arranging n objects where some objects recur. We know that there are n! ways of arranging n distinct objects. So, for example, the number of words (both sensical and nonsensical) that can be made from the letters in this word:

GMAT

is 4!.

What about the number of words that can be created from this word:

DESERT?

Well, we have 6 objects. If they were all distinct, there would be 6! ways of arranging all the letters, and thus 6! words would be made. HOWEVER, not all of the letters are distinct. In particular, "E" shows up twice. So, there are actually 6!/2! words we can create. What about this word:

DESSERT?

Now, we have 7 objects. But both "S" and "E" show up twice. So, there are 7!/(2!*2!) ways of arranging.

DESSERTS?

Now, there are 8!/(3!*2!) ways of arranging.

SSS

Well, there are 3!/3! or 1 way of arranging all the letters.

The formula for arranging n objects where some objects recur is: n!/(r!*s!) in which "r" and "s" are the number of times objects of a certain kind appear.

So, with:

UUURR,

there are 5!/(3!*2!) ways of arranging.

This makes complete sense when we think about just the different permutations available with the various streets and avenues. I couldn't get myself to use this because I wasn't sure if this would totally eliminate the possibility of moving left or down instead of just up and right?

How can we be certain that this wouldn't count the negative distances?
Re: PS: Combinatorics   [#permalink] 23 Apr 2014, 18:25

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