Four dice are thrown simultaneously. Find the probability that two of them show the same face and remaining two show the different faces.

Select a number which ocurs on two dice out of six numbers (1, 2, 3, 4, 5, 6). This can be done in  $C16$, ways.

Now select two distinct number out of remaining 5 numbers which can be done in  $C25$ ways. Thus these 4 numbers can be arranged in 4!/2! ways.

So, the number of ways in which two dice show the same face and the remaining two show different faces is 

  $C16\times C25\times \frac{4!}{2!}=720$

 =>  n(E) = 720

  $\therefore PE=\frac{720}{6}=\frac{5}{9}$