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Let's say I have six balls, three green, two red, one blue. I want to take out two of them and count the permutations:

$$\frac{\frac{n!}{\prod(n_i!)}}{(n-k)!} = \frac{\frac{6!}{3!2!1!)}}{4!} = 10$$

Which is sadly wrong, because the answer is eight. But what is wrong about my function? I know that for k-permutations you need:

$$\frac{n!}{(n-k)!}$$

And for multiset permutation you need:

$$\frac{n!}{\prod(n_i!)}$$

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    The title uses the word "combinations", indicating that order does make a difference, but the body of the Question talks abouts "permutations", which means that order does make a difference. Incidentally your formula for multiset permutation only applies when *all* the items are being arranged (a permutation of $n$ items including some repetitions). – 2017-02-01
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    @hardmath Yes you're right, that's a mistake, I meant to say k-multiset permutation – 2017-02-01
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    Your initial calculation is wrong, which was why the numbers were puzzling me. It works out to $2.5$, which is obviously a massive clue that something is wrong. – 2017-02-01
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    You won't be able to use a simple formula. Instead work out the tree of dependent choices (only two levels are needed ). – 2017-02-01
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    @hardmath Okay I have no idea how to do this – 2017-02-01

3 Answers 3

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Don't overthink this. There is no simple formula. Make a tree of outcomes:

First item (ball) can be any of three colors.

(a) If green, there are three choices for the second colored item.

(b) If red, there are three choices for the second colored item.

(c) If blue, there are (only) two choices for the second colored item.

So the total count of distinguishable outcomes (permutations, taking order into account) is $3+3+2=8$.

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Without getting too deep into how much the form of a calculation "means" something, you can view the top of $\frac {n!}{(n-r)!}$ as saying "consider all possible arrangements" and the bottom as saying "but forget about arrangements of the last $n{-}r$ items".

In your attempted formula, which should calculate out to $\frac {60}{24}=2.5$, you effectively have too many arrangements of the last $4$ items you are "forgetting about", due to identical items where a switch is not a new arrangement. However the real calculation of arrangements for those last four is no easier (or somewhat harder) than directly evaluating for the first $2$, so this formula can't be used for most partial multiset calculations.

In fact you have $3$ choices for the first pick, and then a variable number of choices on the second pick - $3$ or $3$ or $2$, depending on the first pick. As you can see, those alternatives add to the expected answer.

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You are mixing things.

We have simply -

Both Green + Both Red + Both of different Colours

1 + 1 + 6 = 8