Question 1210172
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Given a regular octagon, in how many ways can we color one diagonal red and another diagonal blue 
so that the two colored diagonals intersect at an endpoint? Consider rotations and reflections distinct.
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<pre>
Let's consider any one vertex of the octagon.


The number of diagonals emanating from this vertex is 8 - 3 = 5.

The number of "other" diagonals emanating from this vertex is 5 - 3 - 1 = 4. 


    +--------------------------------------------------------------+
    |       So, there are {{{(5*4)/2}}} = 10 geometrically                |
    |   different pairs of diagonals emanating from each vertex.   |
    +-------------------------------=========----------------------+


Hence, in total, there are 8*10 = 80 geometrically different pairs of diagonals
emanating from 8 vertices of the octagon.


Each pair can be colored in 2 different ways, red-blue or blue-red.


It gives 2*80 = 160 different ways to color all the pairs of diagonals,
emanated from 8 vertices of the octagon. 


<U>ANSWER</U>.  There are 160 different ways to color all the pairs of diagonals,
         emanated from 8 vertices of the octagon. 
</pre>

Solved.


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In my solution, I assume that the question does relate to all possible pairs 
of diagonals for {{{highlight(highlight(all))}}} vertices of the octagon, having a common endpoint at vertices.


I make this addition, since the question in the post is posed in a very unclear way.


If you want the answer for one single vertex, then the number of geometrically different pairs
of diagonals from this vertex is 10 and the number of different coloring of these 10 pairs is 20.


In the post by @CPhill, counting of diagonals in the octagon is made INCORRECTLY.