Simple Combination of balls I encounter this problem but I cannot figure out the solution. Please help.
Given that there are 4 white balls, 2 red balls and 1 black balls of the same size. 4 balls are drawn at random. Find the number of ways this can be done.
The given answer is 1 + 4C1 + 4C1 + 4C2 + 2 × 4! ÷ 2! = 39.
Thank in advanced.

For the answer where balls of the same color are indistinct and order does matter, look above to Dylan's answer.

Assuming all balls are distinct (even those of the same color) and assuming order does not matter
You can break this into cases:
$\begin{array}{|c|c|c||c|}
\hline
\text{White}&\text{Red}&\text{Black}&\text{Total number of cases}\\
\hline
\hline
4&0&0&\binom{4}{4}\cdot\binom{2}{0}\cdot\binom{1}{0}=1\\
3&1&0&\binom{4}{3}\cdot\binom{2}{1}\cdot\binom{1}{0}=8\\
3&0&1&\binom{4}{3}\cdot\binom{2}{0}\cdot\binom{1}{1}=4\\
2&2&0&\binom{4}{2}\cdot\binom{2}{2}\cdot\binom{1}{0}=6\\
2&1&1&\binom{4}{2}\cdot\binom{2}{1}\cdot\binom{1}{1}=12\\
1&2&1&\binom{4}{1}\cdot\binom{2}{2}\cdot\binom{1}{1}=4\\
\hline
\end{array}$
Adding these together gives the answer.  (note: this agrees with the answer $\binom{7}{4}=35$ if we were to simply treat all balls as different and ignore the colors)

Assuming all balls of the same color are indistinct and order does not matter
There will be as many possibilities as there are cases listed above, for a total of 6 cases.
