I know how to find combinations of items where all items are different:
$$P(n,r)=\frac{n!}{(n−r)!}$$
But how do I find it if some items are the same? For example, I have $3$ white and $3$ black elements and want to find out how many different combinations of colors I can make if I arrange them in a row.
And what about $2$ black, $2$ white, $1$ red and $1$ blue element? Or $4$ black, $1$ white and $1$ blue?
If you have $a$ X's, $b$ Y's, and $c$ Z's, then we can start by ordering them ignoring the fact that some are the same:
$$(a+b+c)!$$
And then correct for the distinguishability of each object:
$$\frac{(a+b+c)!}{a!b!c!}$$
(This works when you have any number of objects, not just $3$; you'd simply add $d,e,f...$ to the numerator inside the factorial and divide by $d!e!f!...$)
So for your example with 3 whites and 3 blacks, we'd have $$\frac{(3+3)!}{3!3!} = 20$$ ways.
Can you apply this to your other examples?
(3 white and 3 black) there are $\frac{6!}{3! . 3!}$ ways (2 black, 2 white, 1red, and 1 blue), $\frac{6!}{2! . 2!}$
