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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?

 Dec 8, 2018
 #1
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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. 

 

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

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 Dec 8, 2018

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