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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 10, 2018
 #1
avatar+17286 
+2

1 x 6 = 6

3 x 2 = 6       You have to paint one number from EACH pair

So you have to paint     1  3       1   2       6  3        6 2

4 ways

 Dec 10, 2018
edited by ElectricPavlov  Dec 10, 2018
 #2
avatar+98102 
+1

We have the following number of possible sets of numbers.....we have six numbers and we want to choose any two of them....so....

 

C(6,2) = 15 sets

 

These are 

 

(1,2) (1,3) (1,4) (1, 5) (1,6)

(2, 3) (2,4) (2, 5) (2, 6)

(3,4) (3,5) (3,6)

(4, 5) ( 4, 6)

(5, 6)

 

Note, Ant, that only two of them have a product of six

 

So.....thirteen don't

 

 

cool cool cool

 Dec 11, 2018

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