What is the probability of drawing 1 red pen and 1 green pen? There are 3 blue pens, 2 red pens, 3 green pens and you're drawing two pens at random. What's the probability that 1 will be red and another will be green?
What I tried doing:
$$\frac{\binom{2}{1}\binom{3}{1}\binom{3}{0}}{\binom{8}{2}} = \frac{3}{14}$$
answer says it's $\frac{6}{28}$ though
edit: ok, I'm just stupid didn't know $\frac{6}{28}$ = $\frac{3}{14}$ lol

Consider two slots where the balls will be placed. The total number of events would be $8$ choices for the first slot and $7$ for the second, i.e. $7\times 8 = 56$.
Now, consider your situation. We can get red balls for this slot in $2$ scenarios, and then for the second slot we need to have a green ball whose possibility would be $3$ times, i.e. we would have $6$ of these cases. Similarly, green for the first and red for the second would give us $3\times 2=6$ events.
Therefore, out of 56 times (which are of equal chance, we achieve one red and one green $6+6=12$ times. Thus the probability would be equal to 
$$
\frac{12}{56} = \frac{6}{28}.
$$
