Imagine that you have $25$ identical balls and three boxes, one labelled RED, one labelled BLUE, and one labelled GREEN. Your problem is equivalent to asking how many different ways there are to distribute your $25$ balls amongst the boxes, if you can put at most $15$ balls into the RED box and at most $20$ balls into the BLUE box.
If there were no upper limits, the answer would be $\binom{25+3-1}{3-1}=\binom{27}2$; see here. However, some of these violate the upper limits, so to get the actual answer, we’ll have to subtract those.
First calculate the number of distributions that put too many balls into the RED box. That requires putting $16$ balls into the RED box and then distributing the remaining $9$ balls however you please amongst the three boxes. This can be done in $\binom{9+3-1}{3-1}=\binom{11}2$ ways.
Now calculate the number that put too many (i.e., at least $21$) balls into the BLUE box; the same reasoning shows that there are $\binom{4+3-1}{3-1}=\binom62$ ways to do this.
We don’t have enough balls to put too many balls into more than one box, so the final result is $$\binom{27}2-\binom{11}2-\binom62=351-55-15=281\;.$$
Added: Here’s another way to look at it that may be more familiar to you. Let $x_1,x_2$, and $x_3$ by the number of red, blue, and green balls, respectively, that you select. Then you’re counting integer solutions to $x_1+x_2+x_3=25$ that satisfy the conditions $0\le x_1\le 15$, $0\le x_2\le 20$, and $0\le x_3\le 25$.
Your answer is right. If the selections are distinguished according to the number of red balls and the number of green balls they contain, and all selections except those with $0$ red balls are to be counted, there are $4\times4=16$ different selections.
Best Answer
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.
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: $\frac6{21}=\frac{24}{84}$