counting probability with multiple cases There are four different colors of paint one can use for four different houses. If one color can be used up to three times, how many total possibilities are there? 
I approached the problem by splitting it into cases:
1) each color used once
2) one color used twice
3) two colors used twice
4) one color used thrice
but i dont know how to move on from here or if im right

Choose colours, then choose houses for each colour choice.

1) each color used once

$$\binom{4}{4}\cdot\binom{4}{1}\binom{3}{1}\binom{2}{1}$$

2) one color used twice

(plus two colours used once)
$$\binom{4}{1}\binom{3}{2}\cdot\binom{4}{2}\binom{2}{1}$$

3) two colors used twice

$$\binom{4}{2}\cdot\binom{4}{2}$$

4) one color used thrice

(and one colour used once)
$$\binom{4}{1}\binom{3}{1}\cdot\binom{4}{3}$$

but i dont know how to move on from here or if im right

It's a right approach, but the long one.   It is easier to work with complements.   How many ways can you paint the houses with four colours and not use one colour for all houses.
$$4^4-4$$
