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 $0\le x_3\le 25$.
There are two black balls, and we want to choose one: $\dbinom 21 = 2$.
There are five white balls from which we want to draw two: $\dbinom 52 = \dfrac{5!}{2!\,3!} = \dfrac{5\cdot 4}{2} = 10$..
We use the rule of the product (multiplying) to obtain the total number of ways of choosing one black ball and two white balls:
That gives us $$\binom 21 \cdot \binom 52 = 2\cdot \dfrac{5!}{2!3!} = 2\cdot \dfrac{5\cdot 4}{2} = 20$$
Best Answer
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.