The problem can be solved by counting the number of ways we can fill in the following matrix.

`[;

M=\begin{array}{ccccc} & yellow & red & purple & \sum\ bed & & & & 10\ living & & & & 2\ \sum & 4 & 5 & 3 & \boldsymbol{12} \end{array} ;]`

where [;M_{room,color}\ge 0;] except that the living rooms should be painted different colors. The general problem of counting m x n matrices is not remotely easy. Here, though, you're in luck because you only have to solve three trivial "stars and bars" problems.

• If you paint the living rooms yellow and purple, then the bedrooms have to satisfy y + r + p = 10, where 0 <= y <= 3, 0 <= r <= 5, and 0 <= p <= 2. There is obviously only one solution to this problem, namely, y = 3, r = 5, and p = 2.

• If you paint the living rooms red and purple, then the bedrooms have to satisfy y + r + p = 10, where 0<= y <= 4, 0 <= r <= 4, and 0 <= p <= 2. There is obviously only one solution to this problem, too: y = 4, r = 4, and p = 2.

• If you paint the living rooms yellow and red, then the bedrooms have to satisfy y + r + p = 10, where 0 <= y <= 3, 0 <= r <= 5, and 0 <= p <= 3. This problem is more complicated because you have more paint than you need, but only slightly more complicated because you only have 1 application of paint more than you need. If you use less purple, then there is only one solution: p = 3, y = 3, r = 4. If you use less red, then the solution is r = 4, y = 3, p = 3. And if you use less yellow, then y = 2, r = 5, p = 3.

Each solution excludes the others, therefore there are 1 + 1 + 3 = 5 ways to paint all the rooms.