<s>[INST] Probability of getting Required Sum 
What is the probability of when four dice rolled together once,and getting a sum of Thirteen

If we do by just calculating all possible values of sum, then it will take more time; so we can solve the above problem as Multinomial Coefficents of sum, i.e.:
$$
x_1+x_2+x_3+x_4 = 13\;\text{ where },1\leq x_i \leq 6\;\;\forall\;1\leq i \leq 4.
$$
I got stuck that form the above equation how can we obtain $N_{\text{ways to get $\sum=13$}}$ or is there any easy method to find this kind of problems?
 [/INST] I assume that you already know how to calculate the probability for the sum of two dices. One gets the following table.
$$
\begin{array}{r|c}
k & p(x_1+x_2=k) \\
\hline{} \\
2 & 1/36 \\
3 & 2/36 \\
4 & 3/36 \\
5 & 4/36 \\
6 & 5/36 \\
7 & 6/36 \\
8 & 5/36 \\
9 & 4/36 \\
10 & 3/36 \\
11 & 2/36 \\
12 & 1/36 \\ 
\end{array}
$$
Now we can calculate
$$
p(x_1+x_2+x_3+x_4=13)=$$
$$\begin{array}{}
p( ((x_1+x_2=2) \land (x_3+x_4=11) ) &\lor \\
((x_1+x_2=3) \land (x_3+x_4=10)) &\lor  \\
((x_1+x_2=4) \land (x_3+x_4=9)) & \lor   \\
\cdots &\lor \\
((x_1+x_2=11) \land (x_3+x_4=2))) &= 
\end{array}
$$
$$
\begin{array}{}
p(x_1+x2=2)\cdot p(x_3+x_4=11) &+ \\
 p(x_1+x2=3)\cdot p(x_3+x_4=10) &+ \\
p(x_1+x+2=5)\cdot p(x_3+x_4=9)& + \\
\cdots &+ \\
p(x_1+x2=11)\cdot p(x_3+x_4=2)&=
\end{array}$$
$$\frac{1}{36} \cdot \frac{2}{36} + \frac{2}{36} \cdot \frac{3}{36} + \frac{3}{36} \cdot \frac{4}{36} + \frac{4}{36} \cdot \frac{5}{36} + \frac{5}{36} \cdot \frac{6}{36} + \\ 
\frac{6}{36} \cdot \frac{5}{36} + \frac{5}{36} \cdot \frac{4}{36} + 
\frac{4}{36} \cdot \frac{3}{36} + \frac{3}{36} \cdot \frac{2}{36} + \frac{2}{36} \cdot \frac{1}{36}=$$ 
$$\frac{(2+6+12+20+30 ) \cdot 2}{36 \cdot 36}=\frac{140}{36 \cdot 36 }=\frac{35}{324}$$
</s>