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kchawla
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My approach is by finding the total number of ways to select 3 edges from the 12 total edges of a cube: 12!/(3!9!)=220

Then, we have to subtract out the combination of 3 edges where they meet at the same vertex. This happens at each of the corners of the cube, and there are 8 corners. Thus 220-8=112. Answer B.

Let me know if this is correct.
Small correction: 220 - 8 = 212, not 112, and that extra hundred makes a difference in the answer choices.

- Andrew
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AndrewN Yes, that is right, thank you.
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AndrewN Yes, that is right, thank you.
No worries, ali888. That sort of mistake is easy to make. But I thought it was a shame when you had done the hard work on such a problem. You deserve full credit for your solution. (As an Expert, I feel my role is to provide the community with guidance. I replied to your post with this aim in mind, to serve the community.)

Good luck with your studies.

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This should have GMAT Focus prep tag. ( probably 700-800 range problem )
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sachi-in
This should have GMAT Focus prep tag. ( probably 700-800 range problem )
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Added the tag. Thank you!
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kchawla
For a certain cube, 3 different edges that do not all meet at a single vertex are to be painted red. How many different selections of the 3 different edges to be painted red are possible?

A. 56
B. 112
C. 120
D. 212
E. 504
­Total Edges in a cube = 12
Total combination of 3 edges out of 12 = 12C3 = 12!/(3!*9!) = 220
Total cases where all three edges meet at one point (Unfavorable outcomes) = 8 (at vertices of cube)
Favorable Outcomes = 220-8 = 212

Answer: Option D
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