Three balls are selected at random, what is the number of outcomes? Three balls are selected at random from a bag containing 2 red, 3 green and 4 blue balls.
The balls are not replaced back in the bag after selected.
a) What is the number of outcomes in S?
b) What is the probability of one red, one green and one blue ball?

*

*For a), the answer in my notes is $\binom 93$, however I don't understand why that would be. From how I see it, order here matters. E.g.: If you pick red blue blue or blue blue red, wouldn't that count as 2 separate outcomes?


*Also, by using combinations, aren't we also counting, e.g.: red1 blue1 blue1 and red2 blue1 blue1 as 2 separate outcomes when it should only be one?


*For question b), I simply did $\frac{2}{9} \cdot \frac{3}{8} \cdot \frac{4}{7} = \frac{1}{21}$, however this appears to be wrong as the answer should be $\frac{24}{84}$ which I cannot figure out why.
Thanks.
 A: You can calculate all combinations of red, blue and green selected balls where the sum is $3$. The first line at the table represents the maximal number of balls which are selected.
$$\begin{array}{|c|c|c|} \hline \color{red}2&\color{green}3&\color{blue}4 \\ \hline 0& 0 &3 \\ \hline 0& 1& 2 \\ \hline 0&2 &1 \\ \hline 0& 3&0 \\ \hline1 & 0&2 \\ \hline 1& 1& 1\\ \hline 1&2 &0 \\ \hline 2& 0&1 \\ \hline 2& 1&0 \\ \hline\end{array} $$
Next for every row the number of combinations can be calculated. For instance, the number of selecting $3$ blue balls only is $\binom{2}{0}\cdot \binom{3}{0}\cdot\binom{4}{3}=4$. If you sum the results for every row you´ll obtain $84=\binom{9}{3}$. This example shows that the number of outcomes are $\binom{n}{k}$, where $n$ is the total number of elements and $k$ is the  number of chosen elements.
This is an application of the Vandermonde's identity.

For question b), I simply did 2/9⋅3/8⋅4/7=1/21, however this appears to be
wrong as the answer should be 24/84 which I cannot figure out why.

You have to regard the number of ways. The ways to select one ball for each color are:
$ \color{red}r \color{green}g \color{blue}b, \ \color{red}r \color{blue}b\color{green}g , \ \color{blue}b\color{green}g\color{red}r  , \ \color{blue}b\color{red}r\color{green}g  , \ \color{green}g\color{blue}b\color{red}r,   \ \color{green}g\color{red}r\color{blue}b$
So your result has to be multiplied by 6: $\frac6{21}=\frac{24}{84}$
