An urn contains 4 red, 5 blue and 3 green marbles. Six are drawn at random (without replacement) one at a time. How many ways could the sequence of six marbles contain two of each color?
I thought that it would be
(4c2)(5c2)(3c2)
but that gives me 180 and the answer is 90.
Any help would be highly appreciated!
Best Answer
First, there are $\binom{6}{2}$ ways how to select the red marbles out of the sequence of 6 drawn marbles. Then you have 4 marbles left. You pick 2 of them to be blue, which gives you then $\binom{4}{2}$ possibilities. The remaining marbles are green - so noting to choose. Thus in total you have $\binom{4}{2}\binom{6}{2}=90$ sequences.