Counting the number of ways to paint houses. Problem:

Each of eight houses on a street is painted brown, yellow or white. Each house is painted only one color and each color is used on at least one house. No two colors are used to paint the same number of houses. In how many ways could the eight houses on the street be painted?

Can someone please help me find a way to approach this problem. I don't know where to start.
 A: There must be a color that is used on only one house.  If there were a color having the minimum number of houses that was instead $2$, then there is no way to assign the other colors respecting the constraints of the problem.
Thus the number of houses for the three colors must be either $1,2,5$ or $1,3,4$.
Take the first case $(1,2,5)$:  There are eight houses that could be painted by the first color.  Then there are ${7 \choose 2}$ for the second color, and then the remaining houses must be colored by the third color.
For each of these cases you can permute the colors $3! = 6$ ways, so multiply the above number of ways by $6$.
Now do the analogous for the second case $(1,3,4)$:  Here again there are eight places for the first color, but now ${7 \choose 3}$ for the second, and then the the remaining houses must be colored by the third color.
Again, there are $3! = 6$ alternatives for the choice of colors, so multiply appropriately.
Then add up the two cases.  I get $\boxed{2688}$.
A: Just painting them randomly with out restrictions, we have $3^8$ possibilities ( 3 paint color choices, for 8 houses). That's our sanity check out of the way.
By pigeonhole principle, no matter how we paint them,  we always have at least 1 color, with at least 3 houses. Immediately if we set those aside, we have at most 5 houses remaining to split among the other 2 colors. 
Pigeonhole principle, again tells us in the case of **exactly** 5 houses left, we'll have at least  1 color with at least 3 houses  when split among 2 colors.  But 2 colors with the same number of houses, is forbidden, so at least 1 color has at least 4 houses. Basic arithmetic shows that $4+4+1>8$ so we actually have **exactly** 1 color with at least 4 houses.  
Pigeonhole principle  tells us in the **exactly** 4 houses case, the remaining can't be split without at least 1 color  with at least 2 houses. But 2 colors with the same number of houses, is forbidden, so at least 1 color has at least 3 houses.  This end with a 4,3,1 split ( thankfully). 
Now on to the **exactly** 1 color with **exactly** 5 houses case...
In that case pigeonhole principle says at least 1 color has at least 2 houses . (Thankfully no ties).  5,2,1 split
In the case  of **exactly** 1 color with **exactly** 6 houses, we hit an unbreakable tie, due to our minimum coloring requirement( left it out throughout until now), we can also see we can't have 7, or all 8 houses colored the same.
Now onto the interesting stuff ... 
The first split has $\binom{8}{4}$ for the first selections, $\binom{4}{3}$ for the secondary selections, and the last is forced. This can Also be written as $\binom{8}{4,3,1}=280$ a multinomial coefficient. 
The second split has $\binom{8}{5}$ for the first selections, $\binom{3}{2}$ for the secondary selections, and the last is forced. This can Also be written as $\binom{8}{5,2,1}=168$ another multinomial coefficient. 
Multinomial coefficients, are just a generalization of binomial coefficients. 
Each of the above have 6 ways to choose the colors ( can be thought if as $\binom{3}{1,1,1}=3!=3\cdot 2\cdot 1$ ) , so by distributing over their sum we get $$6\cdot (280+168)=6\cdot 280+ 6\cdot 168= 1680+1008=2688$$
