The only combinations of rolls with sum exceeding 15 are
these 4: 

4+6+6 = 16
5+5+6 = 16
5+6+6 = 17 
6+6+6 = 18

However, each of the first 3 rolls have 3!/2! = 3 distinguishable permutations.
That's 3×3 or 9 ways. The fourth roll 6+6+6 can only be had in one way. So
that's a total of 10 ways the sum can exceed 15.

The number of possible rolls is 6·6·6 = 216

So the desired probability if 10 ways out of 216, or ,
which reduces to , choice B).

Edwin