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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?

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    I would interpret the word "sequence" to mean that order matters, that is, that blue-then-green-then-green is not the same as green-then-green-then-blue.2012-01-19
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    @Gerry: I think that João was taking order into account, since he mentioned a specific arrangement, $GBG$, not just a two-one split, but he overcounted by $2$.2012-01-19
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    If we are counting the number of *strings* of length $3$, we are by definition dealing with permutations, since order matters. But when some of the objects are identical, in counting the number of permutations, we may end up *using* $\binom{n}{k}$, which is first introduced when combinations are discussed.2012-01-19

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