+1206 An ordinary 6-sided die has a number on each face from 1 to 6 (each number appears on one face). How many ways can I paint two faces of a die blue, so that the product of the numbers on the painted faces isn't equal to 6?
+2440 If there is an ordinary six-sided die and you are painting two sides with the same color, then you can calculate the number of ways this can occur.
Therefore, the total number of choices is equal to \(6*5\) or 30. However, the order in which each face is colored is really immaterial in this problem; for example, if I paint the side 4 and then side 1, then it is the same as painting side 1 and then 4. Because of this, I must divide by the number of combinations of duplicates, 2, in this case.
Therefore, there are \(\frac{6*5}{2}=3*5=15 \text{ combinations}\) .
We still are not done. We need to determine which combinations result in the product of a 6. Painting 1 and 6 or 2 and 3 would result in a product of 6. This means there are two combinations that do not meet the original condition.
\(15-2=13\text{ combinations}\)
