If there are three yellow, four black, six red, two green and six purple balls. If one wants five balls, how many ways are there to pick one out of each color?
Spontaneously I think about the multinomial coefficient $\displaystyle\frac{n!}{k_1!k_2!\cdots k_m!}$ Where $k_1$ is, for instance, three yellow, and $k_m$ is four black.
Thus, $\displaystyle\frac{21}{3!4!6!2!6!}$. It does not seem to be right though.
Your given result,
$\displaystyle\frac{21!}{3!4!6!2!6!}$ gives number of possible arrangements of the $21$ balls, when order matters (and when balls of the same color are identical or cannot be distinguished from one another)
Now let's focus on the question. Suppose balls of same color are distinct or marked.
Then, there are $\dbinom{3}{1}$ ways of picking a yellow ball, $\dbinom{4}{1}$ ways of picking a black ball,$\dbinom{6}{1}$ ways of picking a red ball, $\dbinom{2}{1}$ ways of picking a green ball and $\dbinom{6}{1}$ ways of picking a purple ball. Thus our total selection comes to $5$ balls.
Therefore, required ways is
$\dbinom{3}{1} \dbinom{4}{1}\dbinom{6}{1}\dbinom{2}{1}\dbinom{6}{1}$
Suppose balls of same color are identical or they cannot be distinguished from one another.
There there is only one way of picking each color ball.
Therefore total number of ways is $1$
There are $5$ balls and $5$ colors, therefore we are picking exactly one of each color. The number of ways to do this is $3\times4\times6\times2\times6 = 864$. This is because we have $3$ choices for the yellow ball, $4$ choices for the black ball, $6$ choices for the red ball, $2$ choices for the green ball, and $6$ choices for the purple ball. All of these choices are independent, so the total number of possibilities is the number of options at each choice multiplied together.
If the order of the picks matters, then multiply the entire thing by $5!$.
It can't be right because it isn't an integer. You have three choices of yellow balls, four of black, and so on. The choices are independent, so multiply them.