On understanding a combinatorial argument: I am facing some problems in understanding this combinatorial argument:

Number of ways of putting $19$ identical balls in $16$ non-overlapping
  triangles such that no triangle is empty and there is a triangle
  containing exactly $3$ balls = $16 \times 15$

Could anybody rephrase this in way which provides more into why the calculation for this is $16 \times 15$?
ADDED:
What if the balls are distinct,how to approach that problem?  
 A: You have to have one triangle with 3 balls, one with 2 balls, and 14 with 1 ball each.  So you pick the triangle for 3 (16 ways), then pick some other triangle for 2 (15 ways).
A: Every one of the 16 triangles needs at least 1 ball in it, and at least one triangle needs exactly 3 balls in it. By Pigeonhole, this means that one triangle will have three balls, one triangle will have two balls, and the other triangles will all have one ball each. The triangles are distinct so you just have to pick the triangle with the three balls (16 choose 1 choices) and then one for the two balls (15 choose 1 choices) and the rest is determined (14 choose 14 choices). That's 16$\times$15$\times$1 = 240 possibilities in total.

Now suppose the balls are distinct. We could at first consider every permutation of the balls while in the triangles distinct (which would mean 19! times our original 240 possibilities), but permuting any of the balls independently in isolation in either the 3-ball triangle or the 2-ball triangle doesn't effect any change so we divide by $3!$ and $2!$ to get $${16\choose1}{15\choose1}{14\choose14}\frac{19!}{3!2!}=2432902008176640000.$$
