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There is a group of 10 objects, 2 red, 3 blue and 5 green. If the 5 green objects should always be placed together, in how many ways we can put them on a line.

I did this: As 5 places are occupied by the 5 green, that can be disposed in only 1 way as they are indistingushable, I did:

$ 5! / (3! * 2!) = (5 * 4 * 3 * 2 * 1) / *6 * 2) = 120/12=10 $

but I am not sure because the 5 green can be in any space. I tried to draw it on paper and it came that 5 green on 10 spots can be arranged in 6 different ways.

So should I multiply 10 * 6 = 60?

I am not sure thought what is the formula for how to arrange the 5 on 10 spots

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Your answer seems to be correct. $5$ green objects may be placed in $6$ ways, as you correctly found. After that we have $5$ places for $5$ objects of two types. This problem can be reduced to a simple permutation problem: we can just find in how many ways blue or red objects can be places and then fill the rest places with the objects of another color.
Thus, we have $\binom 53$ (if we consider permutations of blue) or $\binom 52$ (if we take red) for every fixed position of green objects. Anyway, $\binom nk = \binom {n}{n-k}$, so we can say that total number of permutations is $6 * \binom 52 = 6 * \frac{5!}{2!3!} = 60.$

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  • $\begingroup$ Thanks Andrei, but how I get to 6 with a formula? because I draw it on paper, but not sure how to get it.. $\endgroup$ – jsab Nov 12 '14 at 13:08
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The easiest way I could see this is:

This is the same task as arranging 6 objects: 2 red, 3 blue and 1 block-of-five-greens.

If you have $a_i$ objects of type $i$, there are $\frac{(\sum{a_i})!}{\prod{(a_i!)}}$ permutations.

Hence you are correct. There are $\frac{6!}{3! 2! 1!} = 60$ arrangements.

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