Number of ways to pick balls with conditions on their colors The problem it's in my native language which isn't English so this is my translation: 
There are 12 balls with different colors (12 colors) and you need to pick 5 balls, which at least 1 is red and you can't have more than 2 balls of same color. On how many ways can you pick the balls (1 by 1). 
Example: Red Red Green Blue Yellow or Green Red Blue Blue Purple
Edit: New translation: You go in a ice cream store. There are 12 different flavors you can pick from. You want to buy ice cream with 5 ice-cream balls/flavors, but you must have(pick) at least one chocolate flavor/ball and you cannot have more than 2 flavors(balls) of the same kind. How many ways there are to build(buy) the ice-cream?
This is my solution which I don't know if its correct:
You must pick at least 1 red ball so it can be on 5 different spots. Then you can pick from all 12 balls and after that you can pick 11 caz red could be in those 12 so it can't be picked again. Again 11 (for the 4th spot) and in the end 10 ways.
So in the end we got 5*12*11*11*10 = 7260 ways the you can pick the balls.
 A: There are $\binom{11}4=330$ ways to choose $4$ non-chocolate flavors and $5!=120$ ways to arrange them and chocolate in a cone with $5$ different flavors; that accounts for $330\cdot120=39600$ different possibilities. 
We could also have a cone with $4$ different flavors, one of them occurring twice. There are $\binom{11}3=165$ ways to choose the $3$ non-chocolate flavors and $4$ ways to choose which flavor will appear twice. Finally, there are $\binom52=10$ ways to decide which two scoops will be the repeated flavor, and there are then $3!=6$ ways to arrange the remaining $3$ scoops. This adds another
$$165\cdot4\cdot10\cdot6=39600$$
possibilities.
Finally, we could have just $3$ different flavors, two of them occurring twice. There are $\binom{11}2=110$ ways to choose the two non-chocolate flavors and $\binom32=3$ ways to choose which two flavors are repeated. Say that the repeated flavors are $A$ and $B$, and the non-repeated flavor is $C$. Then there are $\binom52=10$ ways to choose which two scoops are $A$ and $\binom32=3$ ways to choose which two are $B$, for a total of
$$110\cdot3\cdot10\cdot3=9900$$
possibilities.
The grand total is therefore $39600+39600+9900=89100$ different possibilities.
