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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?

User RDL
by
8.3k points

2 Answers

5 votes

Answer:

13 ways

Explanation:

Total number of distinct ways to pair two faces

= C(6,2) = 6! / (2!4!) = 15

Total number of ways to pair two faces so that the product equals six

=cardinality {1*6, 2*3} = 2

Therefore

number of ways to paint two faces such that the product is not six

= 15 - 2

=13

User Vershov
by
8.7k points
3 votes

Answer:

13

Explanation:

There are 6 ways to choose the first number and 5 ways to choose the second number to paint so the total number of ways is 6 * 5 = 30 but we are over counting by a factor of 2 since 1 and 6 is the same as 6 and 1 so it's 30/2 = 15, not 30. The only choices that have a product of 6 are 2 and 3 or 1 and 6 so the answer is 15 - 2 = 13.

User Heman Gandhi
by
7.7k points

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