GATE Overflow Test Series | Mock GATE | Test 2 | Question: 31

Question

How many ways can you paint the faces of a regular tetrahedron with four colors if each face is painted a different color? (Assume that two paintings that can be oriented to look the same are considered indistinguishable).

• A. $6$
• B. $24$
• C. $2$
• D. None of these

Answer 1

There are $4!$ ways to paint $4$ faces with $4$ different colors, and there are $4 \cdot 3$ orientations of the tetrahedron, so there are only $ \frac{4!}{4 \cdot 3} = 2$ distinct ways to paint the faces.

Comments

@gatecse Sir we can have the above 12 configurations only for this setup right? Now if we take any one of them we can derive the rest 11 by changing the orientations. Isn't it? So only 1 configuration possible. Sir please clarify if I am thinking it in the wrong way.

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you put so much efforts to make others visualize this problem. but i guess all your 12 possibilities are same to one because this all representations can be oriented to look same.

and also you didn’t make other 12 possibilities which all will be same to another one representation. that other representation will be same like your the first one just swap green and yellow in it. (you can swap any two)

its all according to how i can imagine. and could be wrong.

if what i said can help you understand question and answer please explain this question with proper visualization. thanks:)

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@viniit Thank you so much for pointing out where I was going wrong, Yes there will be another 12 possibilities which will again actually comprise of 1 arrangement. I will update the comment. Thanks again.