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Question : When five standard six-sided dice are rolled, what is the probability exactly two different numbers are rolled?

What I did : Since there are five 6-sided dice the total # of possibilities would be $6^5$ or $7776$.

Now this is the part that I am not sure I am right.

Since there are $5$ dice, $\binom{5}{2}$ would give $10$ different possibilities for the $2$ numbers to be the same. Therefore, I would subtract $10$ from the total # of possibilities to get $5/3888$ as the probability.

I am probably wrong so any help would be appreciated.

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    $\begingroup$ There are $\binom 62$ ways to choose the two numbers. Having chosen them there are $2^5$ ways to assign each die (as each has $2$ options). We have to exclude the two cases in which all five die have the same value so there are $\binom 62 \times \left(2^5-2\right)$ ways in the end. $\endgroup$ – lulu Feb 20 '18 at 22:07
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You already correctly determined that the total number of possibilities is $7776$. How many of those have the desired property?:

To produce a configuration we can imagine splitting the $5$ dice into two bunches and then assign a different number to each bunch. Without loss of generality the first die is in the first bunch, there are $2^4-1$ ways of splitting the other dice such that the second bunch is not empty. Then there are $6$ possible values for the first bunch and $5$ possible values for the second bunch. In total this gives $6*5*(2^4-1)=450$.

Another approach is to first decide what the two numbers are going to be, there are $\binom{6}{2}=15$ possible choices. Then we choose a subset (which is not allowed to be empty or everything) of dice for the smaller number; there are $2^5-2$ such subsets. The bigger number automatically gets assigned to all the dice which didn't get the first number. In total $15*(2^5-2)=450$.

The final probability is therefore $450/7776$ which is about $6$ percent

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