Possibilities of rolling 15 with top 3 six-sided dice There are $7776$ possible outcomes of rolling $5$ six-sided dice, of this total there are $651$ possible outcomes where the $5$ dice equal exactly $15$. 
How could you calculate the number of possible outcomes where the top three dice equal $15$?
 A: There are three ways to roll 15 with three dice: 663, 654 and 555. I'll call this the greatest-sum-set (GSS); these have to be the greatest within the roll. I will split each case into three sub-cases, depending on how many times the smallest number in the GSS appears on the remaining two dice:


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*663|xx (x is 1 or 2, the x's can be different). 10 ways to place 6's, 3 ways to place 3 afterwards, the remaining spaces can be filled in 4 ways. Total 10 × 3 × 4 = 120 ways.

*663|3x (one 3 on remaining dice). 10 ways to place 6's, 3 ways to place x, x can have 2 values. Total 10 × 3 × 2 = 60 ways.

*663|33 (two 3's). Only 10 ways to place 6's and no choice after that.

*654|xx (x < 4). 5 × 4 × 3 ways to place 654, remaining spaces can be filled in 9 ways. Total 540 ways.

*654|4x. 5 × 4 × 3 ways to place 65x, x can have 3 values. Total 180 ways.

*654|44. 5 × 4 ways to place 65, no choice afterwards. Total 20 ways.

*555|xx (x < 5). 10 ways to place x's, which can assume 16 values. Total 160 ways.

*555|5x. 5 ways to place x, which can assume 4 values. Total 20 ways.

*555|55. Only one way for this.


Adding up all cases gives 1111 5-dice rolls where the top three numbers sum to 15.
