|
|
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.
RenB
Joined: 13 Jul 2022
Last visit: 30 Apr 2026
Posts: 388
Given Kudos: 304
Location: India
Concentration: Finance, Nonprofit
Schools: Stanford '26 Wharton '26 (D) Booth '26 (D) Yale '26 (D) CBS '26 (A$$$) Darden '26 (WL) HBS '26 (D) Ross '26 (A$$$$) LBS '26 (D) Tuck '26 (A$$) Kellogg '26 (A)
GMAT Focus 1: 715 Q90 V84 DI82 
GPA: 3.74
WE:Corporate Finance (Consulting)
Schools: Stanford '26 Wharton '26 (D) Booth '26 (D) Yale '26 (D) CBS '26 (A$$$) Darden '26 (WL) HBS '26 (D) Ross '26 (A$$$$) LBS '26 (D) Tuck '26 (A$$) Kellogg '26 (A)
GMAT Focus 1: 715 Q90 V84 DI82
Posts: 388
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
61% (02:16) correct
39%
(02:26)
wrong
based on 1087
sessions
History
Date
Time
Result
Not Attempted Yet
Consider a 3 dimensional analog of a rectangular grid with dimensions 2 by 3 by 3. What is the number of paths from the front left-down vertex to the back-right up vertex such that each path consists of travelling only back, right or up?
A. 320
B. 480
C. 560
D. 600
E. 8!
OG Advanced quant- Tip section
(Bunuel or moderator- only the answer was given- kindly edit suitable options as required)
(Bunuel or moderator- only the answer was given- kindly edit suitable options as required)
01 May 2023, 04:23
Kudos
Bookmarks
The image attached with the question is a bit misleading as it is a 2D image.
The 3D path can be visualized as shown below
Screenshot 2023-05-01 164850.jpg [ 11.39 KiB | Viewed 14336 times ]
To move from start to finish, one needs to go 3 units up, 3 units right, and 2 units back. However, the traversal can be in any order. Hence, the entire movement can be visualized as various arrangements of
R R R U U U B B
R ⇒ Right
U ⇒ Up
B ⇒ Back
The number of possible ways =\( \frac{8!}{3! * 3! * 2!}\) = 560
Option C
The 3D path can be visualized as shown below
Attachment:
Screenshot 2023-05-01 164850.jpg [ 11.39 KiB | Viewed 14336 times ]
To move from start to finish, one needs to go 3 units up, 3 units right, and 2 units back. However, the traversal can be in any order. Hence, the entire movement can be visualized as various arrangements of
R R R U U U B B
R ⇒ Right
U ⇒ Up
B ⇒ Back
The number of possible ways =\( \frac{8!}{3! * 3! * 2!}\) = 560
Option C
General Discussion
27 May 2023, 00:09
Kudos
Bookmarks
Hello Bunuel,
I really need your help with this question. I know how to solve the path-based questions but I was not able to make an image by reading this question. I thought that 2 by 3 by 3 are lengths. I never imagine an image with blocks as explained by gmatophobia. What did I miss in this question? how should I have analyzed this question?
Also, The Answer will be:
\(\frac{8!}{3!*3!*2!}\)
why do we divide the 8! with 3!*3!*2!. In combination, we divide like this for repetition, is the same logic working here? if yes, can you explain?
Thank you in advance
I really need your help with this question. I know how to solve the path-based questions but I was not able to make an image by reading this question. I thought that 2 by 3 by 3 are lengths. I never imagine an image with blocks as explained by gmatophobia. What did I miss in this question? how should I have analyzed this question?
Also, The Answer will be:
\(\frac{8!}{3!*3!*2!}\)
why do we divide the 8! with 3!*3!*2!. In combination, we divide like this for repetition, is the same logic working here? if yes, can you explain?
Thank you in advance
