# permutations indistinguishable objects and groups

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

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

• Thanks Andrei, but how I get to 6 with a formula? because I draw it on paper, but not sure how to get it..
– jsab
Nov 12, 2014 at 13:08

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.