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?

Logic Dec 8, 2018

#1**+1 **

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.

- There are six choices for the first face.
- There are five choices for the second face.

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.

\(15-2=13\text{ combinations}\)

.TheXSquaredFactor Dec 8, 2018