Events & Promotions
|
|
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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Jun 0210:00 AM PDT
-11:00 AM PDT
Probability is one of the most important GMAT Quant topics because it often combines logic, counting, set theory, and permutations & combinations. Many students try to solve probability questions by listing every possible case, but GMAT probability...
May 2110:00 AM EDT
-11:59 PM EDT
Make the most of your break with the most realistic GMAT™ prep. Take up to $700 off select products.
May 2910:00 AM IST
-11:00 PM IST
Start your journey with a fully customized action plan and work with a dedicated mentor to achieve a 735+ score.- Jun 10
06:00 AM PDT
-06:15 PM PDT
Register for the GMAT Club Virtual MBA Spotlight Fair – the world’s premier event for serious MBA candidates. This is your chance to hear directly from Admissions Directors at nearly every Top 30 MBA program..
27 Jun 2008, 02:30
Kudos
Bookmarks
N
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
(N/A)
Question Stats:
0% (00:00) correct
0%
(00:00)
wrong
based on 0
sessions
History
Date
Time
Result
Not Attempted Yet
In certain city, a car is parking at A destination, it will go to B destination along routes that are confined to square grid of x streets and y avenues. How many routes from A to B can the car take that have the minimum possible length?
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
27 Jun 2008, 02:44
Kudos
Bookmarks
Minimum length would need to go always right or up. x moves to do "up" and y moves to do "right" i.e. x+y moves total
Among theses x+y moves, we can choose which of these will be the y "right moves", so number of routes that have the minimum length is:
\(C_{x+y}^y=\frac{(x+y)!}{x!y!}\)
Among theses x+y moves, we can choose which of these will be the y "right moves", so number of routes that have the minimum length is:
\(C_{x+y}^y=\frac{(x+y)!}{x!y!}\)
27 Jun 2008, 08:05
Kudos
Bookmarks
Unless Oski is using a notation I'm unfamiliar with, I think he has x+y and y upside down on the left side of the equation above. In any case, it's not quite the correct answer.
No matter how you travel from A to B, if you can only go North or East, you must go North x-1 times (not x times), and you must go East y-1 times (and not y times). The journey will take (x-1) + (y-1) = x+y-2 steps in total.
Once we have these numbers, Oski's explanation of how to find the total number of paths is entirely correct. I tend to think in terms of words: any 'word' we can make with precisely x-1 Ns and y-1 Es will define a path from A to B. For example, if we have 4 streets and 3 avenues (and thus need to go North 3 times and East 2 times), the word: NNNEE defines the path 'go North three times, then East twice'. The word EENNN is a different path: 'go East twice, then North three times'. Each word you can make with exactly two Es and exactly 3 Ns defines a path from A to B. So really, the question is asking: how many 'words' can you make using exactly x-1 Ns and exactly y-1 Es. We have x+y-2 letters to fill in, and can choose the slots for the Es in "(x+y-2) choose (y-1)" ways. Thus the answer should be:
(x+y-2)!/[(x-1)!*(y-1)!]
No matter how you travel from A to B, if you can only go North or East, you must go North x-1 times (not x times), and you must go East y-1 times (and not y times). The journey will take (x-1) + (y-1) = x+y-2 steps in total.
Once we have these numbers, Oski's explanation of how to find the total number of paths is entirely correct. I tend to think in terms of words: any 'word' we can make with precisely x-1 Ns and y-1 Es will define a path from A to B. For example, if we have 4 streets and 3 avenues (and thus need to go North 3 times and East 2 times), the word: NNNEE defines the path 'go North three times, then East twice'. The word EENNN is a different path: 'go East twice, then North three times'. Each word you can make with exactly two Es and exactly 3 Ns defines a path from A to B. So really, the question is asking: how many 'words' can you make using exactly x-1 Ns and exactly y-1 Es. We have x+y-2 letters to fill in, and can choose the slots for the Es in "(x+y-2) choose (y-1)" ways. Thus the answer should be:
(x+y-2)!/[(x-1)!*(y-1)!]
