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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:
62% (02:15) correct
38%
(02:26)
wrong
based on 1167
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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)
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gmatophobia
Quant Chat Moderator
Joined: 22 Dec 2016
Last visit: 09 Sep 2026
Posts: 3,174
Location: India
Concentration: Strategy, Leadership
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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 15121 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
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The 3D path can be visualized as shown below
Attachment:
Screenshot 2023-05-01 164850.jpg [ 11.39 KiB | Viewed 15121 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
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Power of Tiny Gains
1.01^(365) = 37.78
0.99^(365) = 0.03
Power of Tiny Gains
1.01^(365) = 37.78
0.99^(365) = 0.03
General Discussion
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
gauravsirohi93
Hi,
In the question they have given that it is a 3-dimensional analog of a rectangular grid. That means it is a cuboid with dimensions 3*3*2.
Any path will have: R R R U U U B B
Yes, we use the same logic used in combinations.
rak08
Joined: 01 Feb 2025
Last visit: 31 Aug 2026
Posts: 261
Location: India
Schools: NUS '26 ISB '26 INSEAD Aug '26
GPA: 7.14
Schools: NUS '26 ISB '26 INSEAD Aug '26
Posts: 261
would the question have the image? i had a hard time visualizing the object rubrics.
Bunuel gmatophobia JeffTargetTestPrep
Bunuel gmatophobia JeffTargetTestPrep
Posts: 6
Is this part of the Focus Edition Exam? Does this type of question still appear, or is part of the Classic?
aaronmart14
This question is more about combinations than geometry, so yes, such problems can appear on the GMAT Focus Edition. The problem doesn’t require any specialized or advanced geometric knowledge or formulas, it's purely a counting concept.
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