Combinations possible??

The three vertices of the triangle have, by observation, 2 adjacent balls each... thus 2 options each;

$3\quad \times \quad 2$

the total number of balls on each edge leaving aside the vertices is 2013... each has, by observation, 4 options for adjacent balls.

$3\quad \times \quad 2013\quad \times \quad 4$

Counting the total number of balls in such an equilateral triangle is equivalent to solving the sum of the first 2015 natural numbers... since clearly starting from 1 there is an increase of one ball in each row till 2015. (This is the concept of Triangular Numbers ) ...

The number of balls except for the three edges have to be counted. Each of the balls in this number has, by observation, 6 adjacent balls, i.e. 6 options. So this can solved as

$\left\{ \frac { 2015\quad \times \quad 2016 }{ 2 } -\left( 2013\times 3\quad +\quad 3 \right) \right\} \quad \times \quad 6$

Now the Grand Total comes out to be....

$\left[ \left\{ \frac { 2015\times 2016 }{ 2 } -\left( 2013\times 3+3 \right) \right\} \quad \times 6 \right] \quad +\quad \left( 3\times 2013\times 4 \right) \quad +\quad \left( 3\times 2 \right)$

Since this is clearly understood that all cases are being counted twice (since adjacent balls are in pair, and counting is performed from both sides), and since the question only asks to make selections, implying the order does not matter... The solution to the question will be...

$\frac { \left[ \left\{ \frac { 2015\times 2016 }{ 2 } -\left( 2013\times 3+3 \right) \right\} \quad \times 6 \right] \quad +\quad \left( 3\times 2013\times 4 \right) \quad +\quad \left( 3\times 2 \right) }{ 2 } \quad =\quad 6087315$