Question

Sol. [Key Concept: All faces distinct in $n$ rolls has probability $\frac{{}^{6}C_{n}\ n!}{6^{n}}$; take the complement.]

$$\begin{matrix} & P(n) = 1 - \frac{\ ^{6}C_{n}n!}{6^{n}}\ \ ;n \leq 6 \\ & P(2) = 1 - \frac{\ ^{6}C_{2}2!}{6^{2}} = \frac{1}{6} \\ & P(3) = 1 - \frac{\ ^{6}C_{3}3!}{6^{3}} = \frac{4}{9} \\ & P(4) = 1 - \frac{\ ^{6}C_{4}4!}{6^{4}} = \frac{13}{18} \\ & P(6) = 1 - \frac{6!}{6^{6}} = \frac{319}{324} \end{matrix}$$