Counting number of outcomes The question is I have 2 red balls, 2 green balls, and 1 black ball. How many different combinations I can get?
I can only get the answer from listing all possible outcomes, but could someone tell me how to do this in Combinations and Permutations. And I got the answer is 30, is this correct?
 A: If you want to put them in a sequence, where order matters, you have 5 objects so you have $5! = 120$ many orders. 
However, you have two red balls, and these can be in $2! = 2$ many orders, and you cannot tell the difference. So to compensate for the fact that we have two reds, you divide by 2.
The same reasoning also (independently) holds for the two green balls, so there we alos have to divide for $2! = 2$ many equivalent orders.
So in all we have $\frac{5!}{2! \cdot 2!} = 30$ different sequences.
A: We can select two of the five positions in the row for the red balls in $\binom{5}{2}$ ways.  We can select two of the remaining three open positions in the row for the green balls in $\binom{3}{2}$ ways.  The remaining open position must be filled with the black ball.  Hence, the number of ways of arranging two red balls, two green balls, and one black ball in a row is 
$$\binom{5}{2}\binom{3}{2}\binom{1}{1} = \frac{5!}{3!2!} \cdot \frac{3!}{2!1!} \cdot \frac{1!}{1!0!} = \frac{5!}{2!2!1!} = 30$$
