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suppose i have this set { BLUE, BLUE, GREEN}

how many ways can i arrange this, given that {Blue 1, Blue 2, Green} is the same as {Blue 2,Blue 1, Green}

i cant find a reason to call it combination/permutation? how do we deal this dilemma? tnx

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Haven't you studied permutations where some objects are identical ? –  true blue anil Jan 22 '12 at 13:31

2 Answers 2

EDIT: I misread the question as having a set of 3 colors, but allowing duplication! This is a bit silly, and I clearly should have read the question better.

The real answer: let's first place our Blue pieces. There are 3 slots, and we want 2 of them for our Blue pieces. So there are ${3 \choose 2 }= \frac{3!}{2!1!}$ ways to place them. Afterwards, we place our other piece. There is one piece and one slot, so there are $1 \choose 1$ = $1$ way of doing this. Altogether, we have $\frac{3!}{2!1!} \cdot 1 = \frac{3!}{2!}$ ways.

The solution to the problem that wasn't asked is below (as I originally answered)

It's not a single combination or permutation - it's a set of them. There are several ways to do these. For small cases such as this one, it is sometimes easiest just to break out all the cases.

Suppose that there was only one color. How many ways can you choose them? This is possible with 3 different colors, so that's that. What if there were 2 colors? There are $3 \choose 2$ different ways to choose the 2 colors. What if there were 3 colors? Then you add them together.

I don't know what other math you have come across, but there is a funny way to way overdo this with Burnside's Lemma in group theory that happens to come to mind. It is easy, it turns out, to find the number of different colorings as a function of the number of colors (it's a polynomial, even) using a bit of group theory. Maybe if you're interested, I'll edit this later to include that.

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+1 for motivating group theory! Even if OP is negligent, the other users will find it useful if you add the group theoretic solution! –  user21436 Jan 22 '12 at 13:48

There is a quite simple formula :

$$P_{a,b\cdots}(n)=\frac{n!}{a! \cdot b! \cdots}$$

where $P$ is a number of permutations , $n$ is a number of elements and $a,b, \cdots$ are numbers of indentical elements .

In your case :

$$P_{2}(3)=\frac{3!}{2!}=3$$

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