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I'm having trouble with this problem, I know that if they were distinguishable it would simply be $14!$, but I don't know how to approach the problem if there are groups of indistinguishable objects.

A boy has $6$ red, $4$ yellow, and $4$ green marbles. In how many ways can the boy arrange the marbles in a line if the the order of same color marbles doesn't matter?

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    $\begingroup$ HINT: In how many ways can you choose $6$ of the $14$ positions to be filled by the red marbles? Once you’ve done that, in how many ways can you choose where the yellow marbles are to go? $\endgroup$ – Brian M. Scott Apr 30 '20 at 2:10
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There are a total of 14! ways in which to arrange 14 marbles. However in this context, we would be overcounting. To see this we know there are 6! ways to arrange 6 red marbles. Similarly there are 4! ways to arrange the yellow marbles and 4! ways to arrange the green marbles. So we have over counted by a factor of 6!4!4!, as order of the positioning of marbles of the same color does not matter.

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In how many ways can you arrange this word: RRRRRRYYYYGGGG

Let's simplify this: in how many ways can you arrange this word: RRG?

('R', 'R', 'G')
('R', 'G', 'R')
('G', 'R', 'R')

and this one: RRGG:

('G', 'G', 'R', 'R')
('R', 'G', 'G', 'R')
('R', 'G', 'R', 'G')
('R', 'R', 'G', 'G')
('G', 'R', 'R', 'G')
('G', 'R', 'G', 'R')

(harder) and for this: RRGGY:

1 ('Y', 'R', 'G', 'G', 'R')
2 ('R', 'Y', 'G', 'G', 'R')
3 ('Y', 'G', 'R', 'G', 'R')
4 ('R', 'Y', 'G', 'R', 'G')
5 ('R', 'Y', 'R', 'G', 'G')
6 ('G', 'Y', 'G', 'R', 'R')
7 ('Y', 'R', 'R', 'G', 'G')
8 ('Y', 'R', 'G', 'R', 'G')
9 ('G', 'G', 'R', 'Y', 'R')
10 ('G', 'G', 'Y', 'R', 'R')
11 ('R', 'G', 'R', 'Y', 'G')
12 ('R', 'G', 'Y', 'R', 'G')
13 ('Y', 'G', 'G', 'R', 'R')
14 ('R', 'G', 'R', 'G', 'Y')
15 ('R', 'G', 'G', 'R', 'Y')
16 ('G', 'G', 'R', 'R', 'Y')
17 ('G', 'R', 'G', 'Y', 'R')
18 ('G', 'R', 'Y', 'G', 'R')
19 ('R', 'R', 'G', 'G', 'Y')
20 ('G', 'R', 'R', 'Y', 'G')
21 ('G', 'R', 'Y', 'R', 'G')
22 ('R', 'G', 'Y', 'G', 'R')
23 ('R', 'G', 'G', 'Y', 'R')
24 ('G', 'R', 'R', 'G', 'Y')
25 ('G', 'R', 'G', 'R', 'Y')
26 ('G', 'Y', 'R', 'R', 'G')
27 ('Y', 'G', 'R', 'R', 'G')
28 ('R', 'R', 'Y', 'G', 'G')
29 ('R', 'R', 'G', 'Y', 'G')
30 ('G', 'Y', 'R', 'G', 'R')

in conclusion for RRGGY it will be: $\frac{5!}{2!*2!*1!} = 30$

Now you can solve RRRRRRYYYYGGGG

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All 14 marbles can be arranged (permuted) in 14! ways. Many of these arrangements are identical because same colored marbles are indistinguishable from one another. In order to arrive at only those permutations that are distinct, division is required by each permutation relative to red, yellow, and green marbles; which are 6!, 4!, and 4!, respectively. Thus the distinct permutation of these 14 marbles = 14!/(6!)(4!)(4!)

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This is a more algebraic approach, but explains it nonetheless.

In how many ways can you arrange one R and one G? Two of course.

Compare the above operation to the coeffient of ${R.G}$ in this expression:

${(R+G)^2}\\ = R.R + R.G + G.R + G.G\\ = R^2 + 2.R.G + G^2$

Notice that 2 is the number of ways you can arrange one R and one G, and is apparent once you collect the like terms.

Check out the number of ways of arranging two R and one G using

${(R+G)^3\\ = R.R.R + R.R.G + R.G.R + R.G.G + G.R.R + G.R.G + G.G.R + G.G.G\\ = R^3+3.R^2.G+3.R.G^2+G^3}$

It is the coefficient of ${R^2.G}$ and is 3.

Hence your solution is the coefficient of ${R^6.Y^4.G^4}$ in

${(R+G+Y)^{14}}$

Using the multinomial theorem this can be calculated pretty mechanically as

${\frac{14!}{{6!4!4!}}}$

This method can be extended to many more complex scenarios.

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