If three six-sided fair dice are rolled, what is the probability that two dice show one number, and the remaining die shows another number? If three six-sided fair dice are rolled, what is the probability that two dice show one number, and the remaining die shows another number?
I think the total of possible outcomes is $6^3= 216$, but I don't know how to apply the n C r in this case.
 A: Strategy:  Assume the colors of the dice are blue, red, and green to make the dice distinguishable.  To count the favorable cases:


*

*Choose which two dice show the same outcome.

*Choose which number those dice show.

*Choose which of the remaining numbers the other die shows.


Finally, divide by the total number of possible outcomes, which you have correctly calculated. 
A: There are $6\cdot5\cdot4=120$ ways of three different numbers and $6$ ways of three equal numbers. The remaining $216-126=90$ ways are favorable. The probability in question therefore is ${90\over216}={5\over12}$.
A: Hint, when you can't find a magical formula, start by counting the favourite outcomes. There are $6\cdot6\cdot6$ possible outcomes altogether. But favourite ones are:


*

*$d_1=d_2=1$ and $d_3\in\{2,3,4,5,6\}$. $5$ outcomes.

*$d_1=d_2=2$ and $d_3\in\{1,3,4,5,6\}$. $5$ outcomes.

*$...$

*$d_1=d_2=6$ and $d_3\in\{1,2,3,4,5\}$. $5$ outcomes.


Then count for


*

*$d_1=d_3=1$ and $d_2\in\{2,3,4,5,6\}$. $5$ outcomes.

*$d_1=d_3=2$ and $d_2\in\{1,3,4,5,6\}$. $5$ outcomes.

*$...$

*$d_1=d_3=6$ and $d_2\in\{1,2,3,4,5\}$. $5$ outcomes.


And one more ... 
It should be easy to finish the exercise from here.
A: First we need to select those two dies out of the three which show same number which can be done in $^3C_2$ ways. Now for those two dice with same number the third dice can have five different results to satisfy the condition. So the answer will be  $$\frac{^3C_2\times5\times6}{6^3} = \frac{5}{12}$$
