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.
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
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
$$
\binom633^6-\binom622^6=13620.
$$