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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 red, so that the numbers on the red faces don't add up to $7$?

 Mar 28, 2020
edited by helpppp  Mar 28, 2020
 #1
avatar+1970 
+2

Solution:

 

Let's list out the ways that the red faces don't add up to 7. 

1+3

1+2

1+4

1+5

2+3

2+4

 

So, that is 6 ways that this works.

 

Please tell me if I am correct. 

 Mar 28, 2020
edited by CalTheGreat  Mar 28, 2020
 #5
avatar+1970 
+2

The ways to paint the red faces are:

 

1 and 2

1 and 3

1 and 4

1 and 5

2 and 3

2 and 4

3 and 5

3 and 6

2 and 6

4 and 6 

4 and 5

5 and 6

 

So 12 ways. I see my mistake!

 Mar 28, 2020

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