What is the number of distinct arrangement of a collection of 2 white balls and 2 black balls ?
If we explicite the cases and call W the white balls, B the black ones, we have :
WWBB, BBWW, WBWB, BWBW, WBBW, BWWB.
This makes 6 cases.
My question is the following : is my following reason correct : The number of distinct arrangement is : $\frac{A_4^4}{A_2^2A_2^2}=6$
where the denominator both cancels out the cases of permutations of white balls among them, because they are not distinguable, and the cases of permutation of the black balls among them, because they are not distinguable.
I know that my formula makes 6 : but is is just a coincidence, or is my mathematical reasoning correct ?
Your method is correct.
Here is another method: There are four positions to fill. Choose two of the four positions for the black balls. The remaining two positions must be filled with the white balls. Hence, there are $$\binom{4}{2}\binom{2}{2} = \frac{4!}{2!2!} \cdot \frac{2!}{2!0!} = \frac{4!}{2!2!0!} = \frac{4!}{2!2!}$$ distinguishable arrangements of two black and two white balls. The two factors of $2!$ in the denominator represent the number of ways the black balls can be permuted among themselves and the number of ways the white balls can be permuted among themselves without producing an arrangement distinguishable from the given arrangement.