In how many ways can you draw 3 balls? The are $5$ identical white balls and $2$ identical black balls in the box. In how many ways can you draw $1$ black and $2$ white balls?
 A: 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 $$\binom 21 \cdot \binom 52 = 2\cdot \dfrac{5!}{2!3!} = 2\cdot \dfrac{5\cdot 4}{2} = 20$$
A: Assuming you draw three balls from the box without replacement there are 3 options:
WWB, WBW and BWW since the balls are identical we don't care which black ball is which, or which white ball but if we did care we have
$5 \times 4 \times 2 + 5 \times 2 \times 4 + 2 \times 5 \times 4 = 40 + 40 + 40 = 120$
Ways of drawing these balls.
For the answer you have of 20 the questioner may not care about the order in which the balls are drawn only which balls
For the Black ball there are only 2 possibilities and for the White balls there are $5 \times 4 = 20$ ways these can be drawn but we don't care about the order so $\frac{5 \times 4}{2!} = 10$ ways.
Making a total of $2 \times 10 = 20$
It depends on your interpretation of what's meant by the number of ways: Do you only care about which balls or is the order in which they are drawn important.   
A: You choose the black and white balls independently from each set.  For each set of black balls, you have all the choices of white balls, so you multiply.
