Find all School-related info fast with the new School-Specific MBA Forum

It is currently 20 Oct 2014, 14:52

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

Pat will walk from intersection A to intersection B along a

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
1 KUDOS received
Intern
Intern
avatar
Joined: 20 Jul 2008
Posts: 24
Followers: 0

Kudos [?]: 5 [1] , given: 0

Pat will walk from intersection A to intersection B along a [#permalink] New post 03 Aug 2008, 19:22
1
This post received
KUDOS
8
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

61% (02:06) correct 39% (01:06) wrong based on 199 sessions
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

Image
[Reveal] Spoiler: OA

Attachments

File comment: map
PS combinatorics.doc [26.5 KiB]
Downloaded 1665 times

To download please login or register as a user

Kaplan GMAT Prep Discount CodesKnewton GMAT Discount CodesManhattan GMAT Discount Codes
6 KUDOS received
Manager
Manager
avatar
Joined: 15 Jul 2008
Posts: 209
Followers: 3

Kudos [?]: 24 [6] , given: 0

Re: PS: Combinatorics [#permalink] New post 04 Aug 2008, 05:02
6
This post received
KUDOS
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)
Manager
Manager
avatar
Joined: 22 Jul 2008
Posts: 154
Followers: 1

Kudos [?]: 8 [0], given: 0

Re: PS: Combinatorics [#permalink] New post 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.
Manager
Manager
avatar
Joined: 15 Jul 2008
Posts: 209
Followers: 3

Kudos [?]: 24 [0], given: 0

Re: PS: Combinatorics [#permalink] New post 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.
Manager
Manager
avatar
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 [?]: 40 [0], given: 4

Re: PS: Combinatorics [#permalink] New post 08 Jul 2010, 18:55
Bringing up an old topic because it is giving me headaches. :shock: :shock: :shock:

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

So I made a list

Start to the right
RRUUU
RURUU
RUURU
RUUUR

Start with by moving up
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
14 KUDOS received
Kaplan GMAT Instructor
User avatar
Joined: 21 Jun 2010
Posts: 75
Location: Toronto
Followers: 21

Kudos [?]: 106 [14] , given: 2

Re: PS: Combinatorics [#permalink] New post 08 Jul 2010, 20:55
14
This post received
KUDOS
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.

And how about:

DESSERTS?

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

What about this word:

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.
_________________

Kaplan Teacher in Toronto
http://www.kaptest.com/GMAT

Prepare with Kaplan and save $150 on a course!

Image

Kaplan Reviews

Manager
Manager
avatar
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 [?]: 40 [0], given: 4

Re: PS: Combinatorics [#permalink] New post 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 ???
1 KUDOS received
Kaplan GMAT Instructor
User avatar
Joined: 21 Jun 2010
Posts: 75
Location: Toronto
Followers: 21

Kudos [?]: 106 [1] , given: 2

Re: PS: Combinatorics [#permalink] New post 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!

Image

Kaplan Reviews

Manager
Manager
avatar
Joined: 27 Oct 2009
Posts: 152
Location: Montreal
Schools: Harvard, Yale, HEC
Followers: 1

Kudos [?]: 14 [0], given: 18

Re: PS: Combinatorics [#permalink] New post 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?
Expert Post
10 KUDOS received
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 23348
Followers: 3601

Kudos [?]: 28651 [10] , given: 2807

Re: PS: Combinatorics [#permalink] New post 21 Jan 2011, 16:56
10
This post received
KUDOS
Expert's post
2
This post was
BOOKMARKED
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
Q.png [ 25.04 KiB | Viewed 13994 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.

Answer: C.

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
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Manager
Manager
avatar
Joined: 27 Oct 2009
Posts: 152
Location: Montreal
Schools: Harvard, Yale, HEC
Followers: 1

Kudos [?]: 14 [0], given: 18

Re: PS: Combinatorics [#permalink] New post 21 Jan 2011, 17:19
You're the best Bunuel, +1
Manager
Manager
avatar
Joined: 15 Apr 2011
Posts: 71
Followers: 0

Kudos [?]: 13 [0], given: 45

Reviews Badge
Re: Pat will walk from intersection A to intersection B along a [#permalink] New post 13 Apr 2012, 10:11
I'm not able to understand this one. Could someone explain? thanks.
_________________

http://mymbadreamz.blogspot.com

Expert Post
1 KUDOS received
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 23348
Followers: 3601

Kudos [?]: 28651 [1] , given: 2807

Re: Pat will walk from intersection A to intersection B along a [#permalink] New post 13 Apr 2012, 10:18
1
This post received
KUDOS
Expert's post
Manager
Manager
avatar
Joined: 15 Apr 2011
Posts: 71
Followers: 0

Kudos [?]: 13 [0], given: 45

Reviews Badge
Re: Pat will walk from intersection A to intersection B along a [#permalink] New post 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.
_________________

http://mymbadreamz.blogspot.com

Expert Post
1 KUDOS received
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 23348
Followers: 3601

Kudos [?]: 28651 [1] , given: 2807

Re: Pat will walk from intersection A to intersection B along a [#permalink] New post 13 Apr 2012, 10:44
1
This post received
KUDOS
Expert's post
mymbadreamz wrote:
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.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Intern
Intern
avatar
Joined: 21 May 2012
Posts: 11
Location: United States
Concentration: Finance, Real Estate
GMAT 1: Q47 V35
GPA: 3.86
Followers: 0

Kudos [?]: 5 [0], given: 18

Re: Pat will walk from intersection A to intersection B along a [#permalink] New post 29 Jul 2012, 15:35
Bunuel, great example & explanation! Thank you
1 KUDOS received
Director
Director
User avatar
Joined: 22 Mar 2011
Posts: 613
WE: Science (Education)
Followers: 73

Kudos [?]: 527 [1] , given: 43

Re: Pat will walk from intersection A to intersection B along a [#permalink] New post 30 Jul 2012, 02:24
1
This post received
KUDOS
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.

Answer C
_________________

PhD in Applied Mathematics
Love GMAT Quant questions and running.

Director
Director
avatar
Joined: 29 Nov 2012
Posts: 929
Followers: 12

Kudos [?]: 309 [0], given: 543

Re: Pat will walk from intersection A to intersection B along a [#permalink] New post 21 Jan 2013, 01:47
Great explanation Bunuel, thanks a lot.
_________________

Click +1 Kudos if my post helped...

Amazing Free video explanation for all Quant questions from OG 13 and much more http://www.gmatquantum.com/og13th/

GMAT Prep software What if scenarios gmat-prep-software-analysis-and-what-if-scenarios-146146.html

1 KUDOS received
Intern
Intern
avatar
Joined: 09 Jul 2012
Posts: 11
Followers: 0

Kudos [?]: 1 [1] , given: 6

Re: Pat will walk from intersection A to intersection B along a [#permalink] New post 15 Sep 2013, 11:18
1
This post received
KUDOS
2V3H

Total - 5

so, 5!/3!2! = 10

Ans:c
Senior Manager
Senior Manager
avatar
Joined: 15 Aug 2013
Posts: 284
Followers: 0

Kudos [?]: 16 [0], given: 23

Re: PS: Combinatorics [#permalink] New post 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.

And how about:

DESSERTS?

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

What about this word:

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
    Similar topics Author Replies Last post
Similar
Topics:
2 Experts publish their posts in the topic Pat will walk from intersection X to intersection Y along a gmatexam2009 5 14 Nov 2010, 07:55
3 _____ Y |__|__| |__|__| |__|__| x Pat will walk from aaron22197 8 02 Jul 2008, 06:30
1 Pat will walk from intersection X to intersection Y along a prude_sb 3 13 Mar 2007, 22:30
Pat will walk from intersection X to intersection Y along a mrmikec 4 15 Jun 2006, 19:04
OG-Question 316-Page 118-Pat goes from intersection X to Y Pokhran II 4 14 Jun 2005, 13:34
Display posts from previous: Sort by

Pat will walk from intersection A to intersection B along a

  Question banks Downloads My Bookmarks Reviews Important topics  

Go to page    1   2    Next  [ 22 posts ] 



GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.