In a box there are $9$ balls.Three are blue, $2$ are green and $4$ are red. Three of them are randomly drawn, one by one.What are the possible ways to set a sequence of three balls?
To start solving this problem I set a table, where in the header I put the $3$ colors.Then in each row, I wrote the different quantity arragements of the different colors in order to have $3$ balls.
For example, we can have $3$ red balls or we can have $1$ blue ball and $2$ green balls. In the end they must sum $3$ balls.
For me the balls of the same color are undifferentiated between them.So when we have $1$ blue ball and $2$ green balls is indifferent which green ball is on the right side and which green ball is on the left side.They are all the same!
So I applied for a combination. In the given example I choose $1$ position of $3$ to put one blue ball and $2$ positions of $2$ left to put $2$ green balls. I made it in the same way to each of the $9$ row of the table.And at the end I added each row.The result was 21 different ways.
It was a right decision to solve by using combination?