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
 A: If you could paint any house with any color without restriction, there would be $4^4$ colors. 
Now, note that you're looking at all colorings except for the four colorings which are all of one color. So you have $4^4 - 4$ ways to color the houses. 
A: I think your way is ok.  But easier would be to count all the ways to paint the $4$ houses (with no restrictions), and then subtract off the cases you don't want.
A: 1) each color used once


*

*There are $4!=24$ possibilities.


2) one color use twice


*

*Which color is used twice: $4$ possibilities

*Which houses with this color: ${4\choose 2}=6$

*The other two houses: $3\times2=6$

*All in all, $6\times6\times4=144$ possibilities.


3) two colors used twice


*

*Which colors are used: $4\times3=12$

*First pair of houses: ${4\choose2}=6$

*All in all: $6\times6$ and not $12\times6$, since each case is counted twice (do you see why?)
Thus $36$ possibilities.


To show you why cases are counted twice, if you houses are labeled A,B,C,D, and you choose AB for the first pair with one color and CD for the other pair.
Then, you choose colors, say AB:blue, CD:black. But this possibility will happen also when you choose the pair CD first, with color black, and then the pair AB, with color blue. This happens for all choices.
4) one color used thrice


*

*Which color is used thrice: $4$

*Which houses with this color: ${4\choose3}=4$

*Color of the last house: $3$

*All in all: $4\times4\times3=48$


5) one color for all houses (this one is the remaining case)


*

*There are $4$ possibilities (you choose only the color).


You can check that the total is $4^4=256$.
The "favorable cases" are the first four, which account for $252$ possibilities, but as other have noticed, it's easier to compute only the remaining and subtract: $252=256-4$.
A: 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.
$$4^4-4$$
A: How many ways could you color the four different houses without any constraints?
Call this $N$.
Once you have that figured out, consider your one constraint re-stated as:

You are not allowed to use the same color for all four houses.

Note that there are only four cases in which your constraint is violated; i.e., all four houses color 1, all four houses color 2, all four houses color 3, and all four houses color 4.
So the total will be $N - 4$.
