permutations indistinguishable objects and groups

Question

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 indistinguishable, I did:

$ \dfrac{5!}{3!2!} = \dfrac{5 * 4 * 3 * 2 * 1}{6 * 2} = \dfrac{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

Answer 1

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 $6 * \binom 52 = 6 * \frac{5!}{2!3!} = 60.$

Answer 2

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.