In how many ways 10 objects can be colored? (combinatorics) On a table there are 10 different objects that can be colored using 5 different colors
a) How many ways can the 10 objects be colored? $10 \choose 5$
b) How many ways can the 10 objects be colored using all different colors? $5! \cdot 5!$ because I have to order 5 colors in 10 places, but when 5 places are "colored" I have 5 more place to color.
c) How many ways can the 10 objects be colored in such a way that each color is used at most 3 times? any idea?
Are my first two solutions right? what's about the third question? maybe $ 10 \choose 5 $ - $10  \choose 2$?
 A: Those are not correct answers. Your solution before the edit was considering identical objects and used stars and bars. The edited solution does not answer either. The question says different objects. So for the first question, please note there are $5$ color choices for each object and hence the answer should simply be $ \ \displaystyle 5^{10}$ ways.
The second question seeks that all colors must be used (In contrast, in the first question, it is possible that all objects get colored using only one color as an example). So one approach is to use Principle of Inclusion Exclusion.
The answer to the second question should be,
$\displaystyle 5^{10} - {5 \choose 1} \cdot 4^{10} + {5 \choose 2} \cdot 3^{10} - {5 \choose 3} \cdot 2^{10} + {5 \choose 4}$
Alternatively use Stirling Number of the second kind (wiki) which gives you number of ways to arrange $10$ different objects in $5$ identical heaps. We then multiply by $5!$ to assign $5$ different colors to those heaps.
$\displaystyle 5! \cdot {10 \brace 5}$
Here is a hint for the third: Think if each color is used at most $3$ times, how many minimum colors must be used to color all $10$ objects and how many cases do you get? That is one approach.
A: Your first two solutions are wrong.  There are $5$ choices for the color of each of $10$ objects, so the number of choices in all, by the multiplication rules is $5^{10}$.
For the second problem, you can use the principle of inclusion and exclusion.  There are $5^{10}$ ways to color the objects.  We want to subtract the colorings that use only four colors.  There are $\binom54=5$ ways to choose the $4$ colors so this gives $$5^{10}-5\cdot4^{10}$$ ways.  But what about colorings that use only $3$ colors?  We've subtracted them twice at the previous step, once for each possible fourth color, so now we have to add them back in. We have $$5^{10}-5\cdot4^{10}+\binom53\cdot3^{10}$$  You still need to figure out the adjustments that need to be made for colorings that use only $2$ colors or $1$ color.
For the third one, the easiest way that occurs to me is first to figure out all the ways that you can choose no more than $5$ positive integers no larger than $3$ that add up to $10$.  One example is $3,3,2,1,1$.  There are $\binom52$ ways to choose the color that are use $3$ times, and then $3$ ways to choose the color used twice, and the remaining colors are each used once.  Then there are $$\binom{10}{3,3,2,1,1}$$ ways to use the paint the houses with these colors so $$\binom52\binom31\binom{10}{3,3,2,1,1}$$ for this color pattern.  Do this for all the patterns (there aren't many) and add them up.
