I am trying to prove the following:
Let us consider 6 balls, 2 reds, 2 yellow and 2 blue. I want to compute the probability of chosen 2 balls either blue or red (BB, RR, BR, RB). I can see two ways. First, as I don't care if the balls are either red or blue, I see them as one group with four balls. $P_1= \frac{\binom{4}{2}}{\binom{6}{2}}$
Second, I can compute the individual probabilities and then add all of them P(BB)+P(BR)+P(RB)+P(RR). \begin{equation} P_2= \frac{\binom{2}{2}}{\binom{6}{2}} +\frac{\binom{2}{1}}{\binom{6}{1}} \cdot \frac{\binom{2}{1}}{\binom{5}{1}}+\frac{\binom{2}{1}}{\binom{6}{1}} \cdot \frac{\binom{2}{1}}{\binom{5}{1}}+\frac{\binom{2}{2}}{\binom{6}{2}} \end{equation}
Is $P_1$ correct? If so, I want to prove that $P_1=P_2$ in a general case, i.e., we have $n$ balls where there are $n_1$ balls are of type one and $n_2$ balls of type 2 and $n_3$ balls of type 3 and we want to know the probability of chosen $r$ balls, where $r$ can be composed of either balls of type one or type two.
Edited: I am searching for an algebraic proof. (Thanks @joriki for pointing it out).