What is the probability of rolling three dice and obtaining a sum between 5 to 10(inclusively)? What is the probability of rolling three dice and obtaining a sum between 5 to 10(inclusively)?
For two dice, it is easy to create a list containing the outcomes of each dice since there are only 36 cases. ({1,1}, {1,2}, {1,3}, {1,4}, ... , {6,5}, {6,6}).
But for three dice, there are $6^3$ cases. Therefore, how can I calculate the probability of the given question without writing down every possible combinations ?
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Most of the theory that you will need is contained in this answer.
There are two additional pieces of theory needed.
$\underline{\text{Additional Theory-1}}$
Consider the following problem:

*

*$x_1 + x_2 + \cdots + x_k \leq n.$


*$x_1, x_2, \cdots, x_k \in \Bbb{Z_{\geq 0}}.$
The easy way to handle this is to create the variable $a$ and change the problem to

*

*$x_1 + x_2 + \cdots + x_k + a = n.$


*$x_1, x_2, \cdots, x_k \in \Bbb{Z_{\geq 0}}.$


*Note that by the constraint $x_1 + x_2 + \cdots + x_k \leq n,$ 
$a$ is forced to (also) be a non-negative integer.
$\underline{\text{Additional Theory-2}}$
The standard method of handling a problem like

*

*$x_1 + x_2 + \cdots + x_k = n$


*$x_1, x_2, \cdots, x_k \in \Bbb{Z_\geq {1}}.$
is via the change of variables: 
$y_i = x_i - 1 ~: ~i \in \{1,2,\cdots,k\}.$
This leads to

*

*$y_1 + y_2 + \cdots + y_k = (n-k)$


*$y_1, y_2, \cdots, y_k \in \Bbb{Z_\geq {0}}.$
Clearly, there is a bijection between the set of solutions to the $x_1,\cdots,x_k$ problem and the set of solutions to the $y_1,y_2, \cdots, y_k$ problem.  Therefore, the two problems have the same number of solutions.
For what it's worth, suppose that you are given

*

*$~n,m \in \Bbb{Z^+} ~: ~n < m.$


*$P$ equals the number of solutions to 
$x_1 + x_2 + \cdots + x_k \leq m.$ 
$x_1, x_2, \cdots, x_k \in \Bbb{Z_{\geq 0}}$.


*$Q$ equals the number of solutions to 
$x_1 + x_2 + \cdots + x_k \leq (n-1).$ 
$x_1, x_2, \cdots, x_k \in \Bbb{Z_{\geq 0}}$.
Then, the number of solutions where $n \leq ~\text{the sum} ~\leq m$ 
will be $P - Q.$

One additional point.  Your posting represents a Probability problem, where the answer can be expressed as
$$\frac{N}{D} ~: ~D = 6^3.$$
Since each of the $216$ rolls are equally likely, each Stars and Bars solution is just as likely to occur as any of the other $215$ possible rolls.  Therefore, the problem reduces to enumerating $N$.
