3 red, 3 green, 3 blue, and 3 orange balls are in a box and 6 of these balls are drawn at random from the box. If 2 of the 6 drawn balls are red and 2 of them are green, what is the probability that the other 2 drawn balls are blue and orange?
The answer I am getting is 12/28. I came to this answer by realizing there are 8 balls left. 3 orange, 3 blue, 1 green and 1 red. Two more balls need to be chosen. So the total number of ways these can be picked are $8\choose 2$=28 ways to choose 2 balls from the remaining 8. Then I got that there are 3!+3! ways to pick a orange and a blue ball from the remaining two balls. So i got my answer to be $(3!+3!)/28$=12/18 as the probability of picking a orange and a blue as the other two of the 6 balls drawn. I don't know if this is right. Can anyone confirm my solution. If it is incorrect, could anyone point out where my error lies?