The probability of getting 10 in a single throw of three fair dice...

Total outcomes $=216$ Now, number of cases of getting 10 from 3 dices in single throw are Case $1: 1+3+6 \rightarrow$ outcomes $=3!=6$ Case $2: 1+4+5 \rightarrow$ outcomes $=3$ ! $=6$ Case $3: 2+2+6 \rightarrow$ outcomes =$\frac{3!}{2!}=$3$Case$4: 2+3+5 \rightarrow$outcomes$=3!=6$Case$5: 2+4+4 \rightarrow$outcomes =\frac{3!}{2!}=3$ Case $6: 3+3+4 \rightarrow$ outcomes =$\frac{3!}{2!}=$3$Favourable outcomes$=27\therefore$Probability =\frac{27}{216}=\frac{1}{8}$$\begin{aligned} \Rightarrow & & (x+24)(x-20) & =0 \\ \Rightarrow & & x & =20 \end{aligned}$$$