Three different dice are rolled simultaneously, three times. The probability that all of them show different numbers only two times, is equal to

$\frac{100}{243}$

Probability of showing different numbers on a single trial $=\frac{\left({ }^6C_3\right)(3 !)}{6^3}=\frac{5}{9}$

Thus, required probability

$={ }^3 C_2 .\left(\frac{5}{9}\right)^2 . \frac{4}{9}=\frac{100}{243}$