# Permutation or Combination?

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. –  Gerry Myerson Jan 19 '12 at 11:49
@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 $2$. –  Brian M. Scott Jan 19 '12 at 12:05
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. –  André Nicolas Jan 19 '12 at 17:04

There are three positions and three colours so there are at most $3^3=27$ ways of colouring the positions.

But there are only two green balls, so colouring all three positions green is impossible. The other 26 are possible, so there are $26$ ways of colouring the positions with balls.

They are not equally likely if you draw the balls at random. To do that calculation you would need to consider the balls as individually distinguishable and there would be $9 \times 8 \times 7 = 504$ possibilities.

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My Math book consider the balls as individually distinguishable and therefor, there are 504 possibilities.But how a math problem interpretation can be so subjective in order to ended in two different solutions? –  João Jan 19 '12 at 19:08

The following analysis is correct in spirit but wrong in detail, since I misread the problem slightly; the correct analysis is at the bottom.

It’s true that balls of the same color are indistinguishable, but the sequence of colors does matter. If you had at least $3$ balls of each color, you could get every possible sequence of the three colors: $BBB,BBG,RBB,GRG,GGR,GRB$, etc., and there are $3^3=27$ such sequences. However, you have only one blue ball and two green balls, so some of these sequences are impossible. To get the final result you can either count the possible sequences directly or count the impossible sequences and subtract from $27$. It’s not hard to count the possible sequences directly.

There is one way to have three red balls, $RRR$, and it’s impossible to have three green or three blue balls. Thus, there’s just one way to have all three balls the same color.

Now let’s use exactly two colors. There are three ways to have two red balls and one green ball: $RRG,RGR,GRR$. Similarly, there are three ways to have two red balls and one blue ball, three ways to have two green balls and one red ball, and three ways to have two green balls and one blue ball. We can’t have two blue balls, so we’ve covered the two-color possiblities: there are $12$ of them.

Finally, if we use all three colors, we have one ball of each color, and there are $3!=6$ different arrangements. I get a total of $1+12+6=19$ possible sequences.

As a check, let’s make sure that there are $27-19=8$ impossible sequences. There are two one-color sequences that are impossible, $BBB$ and $GGG$. The only other impossible sequences are those requiring two blue balls. There are three that use a green ball with the two blue balls: $BBG,BGB,GBB$. There are another three that replace the green ball with a red ball. Thus, we do indeed have a total of $2+3+3=8$ impossible sequences.

Added: As Henry points out, I somehow misread the problem. Since there are in fact three blue balls, the only impossible sequence of colors is $GGG$, and the other $27-1=26$ sequences are all possible. I’ll leave the incorrect analysis, since it may be useful to someone attempting a similar problem.

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There are three blue balls –  Henry Jan 19 '12 at 12:13
@Henry: Thanks: somehow I misread it as one. –  Brian M. Scott Jan 19 '12 at 12:18
$1.$If sequence contains balls of different colors then you have $6$ possible arrangements .
$2.$If sequence contains balls of the same color then you have $2$ possible arrangements .
$3.$If sequence contains two balls of the same color then you have $18$ possible arrangements .
So if you sum up them all you will get $26$ possible arrangements .