|A) 4/9||B) 5/9|
|C) 11/18||D) 7/9|
Select a number which ocurs on two dice out of six numbers (1, 2, 3, 4, 5, 6). This can be done in , ways.
Now select two distinct number out of remaining 5 numbers which can be done in 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
=> n(E) = 720