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.
 A: 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.
A: We can consider the different cases that we get depending on how many white balls are selected.
If all $4$ selected balls are white, then there is only one (distinguishable) possibility for the four selected balls: $WWWW$.
If $3$ of the balls are white, then we get two cases depending on whether the remaining ball is red or black. In each case, there are $\binom{4}{1}$ ways to select in which position the non-white ball is, so this is where the two $\binom{4}{1}$ terms in the answer comes from.
If $2$ of the balls are white, then either both remaining balls are red, or there is one red and one black ball.
So if $2$ of the balls are white, and $2$ are red, then there are $\binom{4}{2}$ ways to select which two of the chosen balls were the white ones.
Finally, the remaining cases are all where there are two balls of one colour, and one each of the remaining colours. (Either two white balls, and one red and black, or two red balls, and one black and white)
In each of these cases, there are $\frac{4!}{2!}$ ways in which the balls could have been selected. This is because we can arrange all $4$ balls in $4!$ ways. But two of the balls are indistinguishable, so we have counted each valid arrangement $2!$ times: once for each possible order of the two indistinguishable balls.
This gives us a further $2\cdot\frac{4!}{2!}$ possible arrangements, which makes the total
$$1 + \binom{4}{1} + \binom{4}{1} + \binom{4}{2} + 2\cdot\frac{4!}{2!}=39$$
