In a bag there are $3$ black balls and $9$ white balls. All the balls are drawn one by one without replacement. What is the probability that the first and the second balls have different colors?
In this kind of problem, I always assume that the balls of the same color (and shape) are indistinguishable between them. So, in this problem, I applied this strategy:
Possible cases: $\frac{12!}{3! \cdot 9!}$
Favorable cases:
We can have BW or WB. For each of these situations the other $10$ balls can be chosen in $\frac{10!}{2! \cdot 8!}$ ways.
In the end, the probability is equal to $\frac{9}{22}$. The book's solution is the same; however, the book's strategy considers each ball as individually unique, despite the color or the shape.
What would you do? Thanks.