There are $a$ balls of color A, $b$ balls of color B, $c$ balls of color C in the bag and $d$ balls of color D in a bag. What are the chances that when I take four balls out of the bag at the same time, I take out one of each color?
So I realised that there were $^{a+b+c+d}C_{4}$ ways of picking them. I also realised that there were $4!$ permutations of picking. How go from here?
On
So you are going to put $^{a+b+c+d}C_{4}=\binom{a+b+c+d}{4}$ at the denominator. Then, as regards the numerator of the fraction which gives the probability, consider the product $abcd$: we have $a$ ways to pick a ball of color A, $b$ ways to pick a ball ball of color B...
Combinations is the most natural approach.
By the multiplication rule, there are $$ {\small{\binom{a}{1}}} {\small{\binom{b}{1}}} {\small{\binom{c}{1}}} {\small{\binom{d}{1}}} =abcd $$ sets of $4$ balls, with one from each of $A,B,C,D$.
Since the number of possible sets of $4$ balls is $${\small{\binom{a+b+c+d}{4}}}$$ it follows that the required probability is $$ \frac { abcd } { {\large{\binom{a+b+c+d}{4}}} } $$ Notice that this approach would work just as well if you were asked, for example, to find the probability of selecting $10$ balls with
assuming $a+b+c+d\ge 10$.
Following the same approach, the required probability would be $$ \frac { \binom{a}{1}\binom{b}{2}\binom{c}{3}\binom{d}{4} } { {\large{\binom{a+b+c+d}{10}}} } $$