Six dice are thrown simultaneously. The probability that exactly three of them show the same face and remaining three show different faces, is

$\frac{(5!)^2}{2(6^6)}$

Six dice when thrown simultaneously can result in $6^6$ ways.

T∴ otal number of elementary events = $6^6$.

Select a number which occurs on three dice out of six numbers 1, 2, 3, 4, 5, 6. This can be done in ${^6C}_1$ ways. Now, select three numbers out of the remaining 5 numbers. This can be done in ${^5C}_3$ ways. Now, we have 6 numbers like 1, 2, 3, 4, 4, 4; 2,3,6,1, 1, 1 etc. These digits can be arranged in $\frac{6!}{3!}$ways.

So, the number which ways in which three dice show the same face and the remaining three show distinct faces is

${^6C}_1 × {^5C}_3 ×\frac{6!}{3!}$

∴ Favourable number of elementary events $= {^6C}_1 × {^5C}_3 ×\frac{6!}{3!}$

Hence, required probability $=\frac{^6C_1 × {^5C}_3 ×\frac{6!}{3!}}{6^6}= \frac{(5!)^2}{2(6^6)}$