In how many ways can $3$ balls be selected from the box if at least one black ball is to be included in the draw? 
A box contains $2$ identical white balls, $3$ identical black balls and $4$ identical red balls. In how many ways can $3$ balls be selected from the box if at least $1$ black ball is to be included in the draw?


My try:
If at least $1$ black ball is to be included in the draw, then either $1$ black, $2$ non-black balls can be selected or $2$ black, $1$ non-black balls can be selected or $3$ black, $0$ non-black balls can be selected. This can be done in $$\binom{3}{1}\times\binom{6}{2}+\binom{3}{2}\times\binom{6}{1}+\binom{3}{3}\times\binom{6}{0}=64$$
But the answer given is $6$. I don't know where have I gone wrong. 
 A: You have not taken into account the fact that balls of the same color are identical.  What matters here is how many balls of each color are selected.  
If $b$ is the number of black balls, $r$ is the number of red balls, and $w$ is the number of white balls, then 
$$b + r + w = 3 \tag{1}$$
Since at least one black ball is selected, $b \geq 1$.  Let $b' = b - 1$.  Then $b'$ is a non-negative integer.  Substituting $b' + 1$ for $b$ in equation 1 yields 
\begin{align*}
b' + 1 + r + w & = 3\\
b' + r + w & = 2 \tag{2}
\end{align*}
Equation 2 is an equation in the non-negative integers.  Since there are two white balls, three red balls, and four red balls, there are at least two balls of each color remaining to be distributed.  A particular solution of equation 2 in the non-negative integers corresponds to the placement of two addition signs in a row of two ones.  For instance, 
$$ + 1 + 1$$
corresponds to the solution $b' = 0$, $r = 1$, and $w = 1$, while 
$$+ 1 1 +$$
corresponds to the solution $b' = 0$, $r = 2$, and $w = 0$.  Thus, the number of solutions of equation 2 in the non-negative integers is 
$$\binom{2 + 2}{2} = \binom{4}{2} = 6$$
since we must choose which two of the four symbols (two ones and two addition signs) will be addition signs.
A: Use stars and bars: remove 1 ball (black), you have 2w, 2b, 3r. You need to select 2 balls from 3 sets, i.e. $\binom{2+3-1}{3-1} = 6$
A: As noted by N. F. Taussig, everything here hinges on the interpretation of the word "identical". There are thus two interpretations of the question. 
Consider the simpler question: How many ways can we choose $1$ ball from a bag containing two identical balls?
The two interpretations yield the answers: 1 way (conventional interpretation); 2 ways (your interpretation).
It seems to me that if we take the question literally, there is a contradiction here: if there are two balls, they cannot be "identical". For example, after we take one ball from the bag, we have one ball in the bag, and the other ball not in the bag. Since the two balls do not agree as to whether they are in the bag, they are not identical (they do not share exactly the same properties and are thus distinguishable). 
This seems to be the implicit reasoning behind your interpretation: "if there are 2 balls, they really are distinct even if we can't tell the difference between them." We could say that you count the number of "objectively" different ways the balls can be chosen. Your interpretation is therefore a "correct" one- in the sense that it avoids contradiction. Furthermore your answer is correct with respect to this interpretation.
On the other hand, the conventional  interpretation would seem to succumb to the contradiction mentioned above. However, this is not necessarily the case. We can interpret the conventional interpretation as saying: "if we cannot tell the difference between 2 ways of choosing, then $\textit{as far as we are concerned}$ they the same- we only count the number of ways of choosing balls that an observer could distinguish between." Thus, the conventional (subjective) interpretation is also a correct one. In fact, it is $\textit{the}$ correct interpretation- in the sense that this is what is intended by the question- and the answer $6$  is $\textit{the}$ correct answer.
To summarise, you only "went wrong" by not knowing the convention. Your answer is perfectly reasonable.
