Three ordinary dice are thrown and the points obtained are added up. What is the probability of obtaining 10 points?

Question

Three ordinary dice are thrown and the points obtained are added up. What is the probability of obtaining 10 points?

I tried to build by extension, the set of favorable results:

{

(1,6,3);(1,3,6);(3,1,6);
(2,5,3);(2,3,5);(3,2,5);
(3,4,3);(3,3,4);(3,3,4);
(4,3,3);(4,3,3);(3,4,3);
(5,2,3);(5,3,2);(3,5,2);
(6,1,3);(6,3,1);(3,6,1)

}

They are 18, except the 2 repeated.

Is there a way to do the calculation for these types of exercises, without having to count one by one?

Answer 1

Positive integral solutions of $x_1+x_2+x_3=10$
$=\dbinom{10-1}{3-1}$

Positive integral solutions of $x_1+x_2+x_3=10$ where $x_1\gt6$
$=\dbinom{4-1}{3-1}$
(this does not include $x_2\gt6$ and/or $x_3\gt6$ because if $x_1\gt6,x_2,x_3\lt6$

Positive integral solutions of $x_1+x_2+x_3=10$ where $x_2\gt6$
$=\dbinom{4-1}{3-1}$

Positive integral solutions of $x_1+x_2+x_3=10$ where $x_3\gt6$
$=\dbinom{4-1}{3-1}$

Therefore, required count

$=\dbinom{10-1}{3-1}-3\times \dbinom{4-1}{3-1}=27$