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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 \(4\)?

 Mar 14, 2022
 #1
avatar+36417 
+2

you cannot paint the 1 and the 4  as the two faces

there are    6 c 2   ways to paint two faces = 15 ways

    and ONE way (1 and 4) that you cannot paint    = 14 ways

 Mar 14, 2022
 #2
avatar+1382 
+1

You forgot 2 and 2 EP, thus, there are 13 working cases out of 15 total cases. 

 Mar 14, 2022
 #3
avatar+23183 
+2

I agree with EP on this problem; there is no way to paint 2 and 2 because

there is only one 2.

Those that work:

   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

The one that doesn't:  1,4

 Mar 14, 2022
 #4
avatar+1382 
+2

Ah... I forgot that you roll ONLY 1 dice. Sorry... ūü§¶

BuilderBoi  Mar 14, 2022

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