Bunuel wrote:
Eight congruent equilateral triangles, each of a different color, are used to construct a regular octahedron. How many distinguishable ways are there to construct the octahedron? (Two colored octahedrons are distinguishable if neither can be rotated to look just like the other.)
(A) 210
(B) 560
(C) 840
(D) 1260
(E)}1680
An octahedron has 8 triangles.
So, we have to fix 2 faces, before applying colours.
So, the first face that can be fixed can be of any of the 8 colors but each face will be similar , so 8/8 ways.
Now, when you have fixed one face, the three adjoining faces are similar, so we have 7 colours for the second face, so 7/3.Now, you can place any color anywhere, it will be a different arrangement, so 6!
Total = \((\frac{8}{8})*(\frac{7}{3})*6!=1*7*6*5*4*2=1680\)
E
Edit: To further expand on bold portion above.Look at the sketch.
We can pick any face and color with any of the available 8 colours. Hence \(\frac{8}{8}\), only one way
Now, we have 7 colours for the second face, but the face has to be adjoining to the blue, so that can fix the octahedron for the remaining 6 colors.
There are three adjoining faces, and 7 colors. The colour picked up can be placed in any of the three faces, but each of these will give us same arrangements of colours.
Here we have taken light brown, but it could be any of the 7 ways. Now the placement of this light brown colour can be at any of the adjoining triangle that is sharing a common side with the blue triangle.
Each of the three faces chosen will result in same arrangement and would depend on arrangements of colour for remaining 6 faces.
Example: We paint 1 next. Say light brown and next 2 as yellow and 3 as black and others in say A, B, C, D, E, F colors. Total ways (8/8)*7*6!
However this will include repetitions THREE times for each as shown in the sketch. Hence divided total by 3 = \((\frac{8}{8})*(\frac{7}{3})*6!=1*7*6*5*4*2=1680\)
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