<s>[INST] Counting Principle - Dependent Events There are 7 red and 5 yellow fish in an aquarium. Three fish are randomly caught in a net. Find the probability that the fish were:
a)  All red
b)  Not all of the same color.
I have solved this question using the method for dependent events as follows:
a)  $\frac{7}{12} \cdot \frac{6}{11} \cdot \frac{5}{10} = \frac{210}{1320}$
b)  $1 - \left(\frac{7}{12} \cdot \frac{6}{11} \cdot \frac{5}{10} + \frac{5}{12} \cdot \frac{4}{11} \cdot \frac{3}{10}\right) = \frac{1170}{1320}$
Is it possible to use combinations to solve this question? If yes, how?
 [/INST] $(a)$
$$\frac{7 \choose 3}{12 \choose 3}$$
You can show that this equals what you did.
$$\begin{align*}
\frac{7 \choose 3}{12 \choose 3}
&=\frac{\frac{7!}{3!4!}}{\frac{12!}{3!9!}}\\\\
&=\frac{7\cdot6\cdot5}{\color{red}{3\cdot2\cdot1}}\frac{\color{red}{3\cdot2\cdot1}}{12\cdot11\cdot10}\\\\
&=\frac{7}{12}\cdot\frac{6}{11}\cdot\frac{5}{10}
\end{align*}$$
$(b)$
$$1-\frac{7 \choose 3}{12 \choose 3}-\frac{5 \choose 3}{12 \choose 3}$$
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