How many ways are there of rolling 6 different coloured 6-sided dice so that exactly 3 different values are showing? Any help with this problem would be greatly appreciated. I have tried solving this problem like this:
1st die: 6 choices,
2nd die: 5 choices,
3rd die: 4 choices,
4th die: 3 choices,
5th die: 3 choices,
6th die: 3 choices
But this isn't getting me anywhere! Thanks.
 A: Basically, you need to choose 3 distinct #s from 6,
choose patterns of 4-1-1 of a kind, 3-2-1 of a kind and 2-2-2 of a kind and permute them.
$4-1-1$ of a kind: $\binom31\binom63\cdot\frac{6!}{4!} = 1800$
$3-2-1$ of a kind: $\binom31\binom21\binom63\cdot\frac{6!}{3!2!} =7200$
$2-2-2$ of a kind: $\binom63\cdot\frac{6!}{2!2!2!} = 1800$
Add up to get a total of $10800$ ways  
A: Do this problem in three steps:


*

*Choose 3 of the 6 values

*Partition the six dice into 3 non-empty groups

*Assign the 3 values from step 1 to the 3 groups from step 2
The total number of ways to complete these steps is ${6\choose 3}\times {6\brace 3}\times 3!=20\times 90\times 6=10800.$
A: Hint: count the number of ways of choosing the three values. Once the three values are chosen, you can find the number of ways of having only those values on the six dice. But then you have to eliminate the cases where only two numbers show (and avoid double-counting the one-number-only cases). Work systematically and put the pieces together.


 Since this is now quite an old question, choose $3$ values in $\binom 63=20$ ways. For each of these ways there are $3^6=27^2=729$ ways of obtaining only those values on the six dice. However you might get only two values - there are three two item subsets of the original values, and for each of these we obtain $2^6=64$ ways of scoring just the two values - $3\times 64=192$. Then we might have just one value - the same on all six dice - that can be done in three ways once we've chosen the three values - one way for each value. We counted those in when we did $3^6$, and counted them out again when we did $2^6$ - but we took off more than one $2^6$ and in fact deducted the cases with just one value twice, so we need to add $3$. Then the calculation comes out at $20\times (729-192+3)=20\times 540 =10,800$

The method here is inclusion/exclusion.
A: There are $\binom633^6$ ways with at most three distinct values ($\binom 63$ to pick the three values and $3^6$ ways of assigning values to the dice). Similarly, there are $\binom622^6$ ways with at most 2 values. So the answer is
$$
\binom633^6-\binom622^6=13620.
$$
