There are 5 cubes, each cube has a different color and on each cube the numbers 1-6 There are 5 cubes, each cube has a different color and on each cube the numbers 1-6. Someone throws the cubes.
a. How many results are there? I was thinking: $6^{5}$ ways for the throws regardless of color, multiply by $5!$ ways to order the colored cubes $=6^{5}\cdot5!$
b. In which at least one cube shows the number '3'? I was thinking: let the first cube show the number 3, there are $5$ ways for that. multiply by the rest of the results $=5\cdot6^{4}\cdot4!$
c. In which exactly one cube shows  '2' and exactly one cube shows '4'? I was thinking: let the first cube show '2' and the second '4', there are $5\cdot4$ ways for that. Let the first cube show '4' and the second '2', there are $5\cdot4$ ways for that as well. Then we have $6^{3}\cdot3!$ ways for the rest of the results. So the answer $=2\cdot5\cdot4\cdot6^{3}\cdot3!$
d. In which the set of the numbers that appear on the cubes has exactly 3 objects? I was thinking: we need 3 different numbers and then 2 numbers that appeared already, so - $6\cdot5\cdot4\cdot1\cdot1$, then multiply by all the ways to order the colors ... so the answer $=5!\cdot6\cdot5\cdot4$If someone can help me understand where I was right and where I was wrong I would highly appreciate it!
 A: a) Throws the cubes probably means all at once. One way to record what happened in to list the outcomes on the various dice, in say alphabetical order of colours.
b) There are $5^5$ outcomes in which we don't get any $3$, for the same reason there are $6^5$ outcomes in total. So there are $6^5-5^5$ outcomes in which we get at least one $3$.
c) The die on which we get the $2$ can be picked in $5$ ways. For each choice, the die on which we get the $4$ can  picked in $4$ ways. For each such choice, the results on the $4$ remaining dice can be picked in $4^4$ ways, anything but a $2$ or a $4$ on each. This gives a total of $5\cdot 4\cdot 4^4$.
d) This one is slightly tricky. We use a slightly fancy device called Inclusion/Exclusion. It also could be done more crudely by dividing into cases. 
The set of numbers that appear can be chosen in $\binom{6}{3}$ ways. We count the number of ways one particular such set, say $\{1,2,3\}$ could appear.
There are $3^5$ possible outcomes in which what appears is restricted to $\{1,2,3\}$. But some of these outcomes don't give us all of $1,2,3$. For example, $2^5$ outcomes miss $1$, with the same number missing $2$ and missing $3$. So our next estimate of the answer is $3^5-3\cdot 2^5$. But we have subtracted one too many times each outcome which yields a single number. So the number is $3^5-3\cdot 2^5+3$. To get the total number of possibilities, multiply by $\binom{6}{3}$. We end up with a total of $\binom{6}{3}\left( 3^5-3\cdot 2^5+3   \right)$.
If you wish, and if there is time, I can sketch the cruder cases approach.
