How many different arrangements of $2$ blue, $2$ pink and $2$ green houses are there?

Question

There are $6$ houses, $2$ blue, $2$ pink and $2$ green, how many different arrangement of these houses are there?

My answer is $6\cdot5\cdot4\cdot3\cdot2 = 720$ and $3\cdot2 = 6$. Divide $720$ by $6 =120$... is this correct?

Answer 1

More general if you have a set $S$ of $n$ elements with only $m$ diverse elements and so some of them occur $k_i$-times the total amount of possible arrangements is given by

$$P_n^m=\frac{n!}{k_1!\cdot k_2!\cdots k_m!}$$ with $k_1+k_2+\cdots+k_m=n$

Which is in your case $n=6$, $m=3$ and $k_i=2$ and plugging this in leads to

$$P_6^3=\frac{6!}{2!\cdot 2!\cdot 2!}=90$$

Answer 2

Not quite.

I'm not sure how one is supposed to arrange houses, so let's say we have a row of six houses, and we wish to paint two of them blue, two of them green, and two of them pink.

We choose two of the six houses to paint blue, two of the remaining four houses to paint green, and paint the remaining two houses pink, which we can do in $$\binom{6}{2}\binom{4}{2} = \frac{6!}{2!4!} \cdot \frac{4!}{2!2!} = \frac{6!}{2!2!2!} = \frac{720}{8} = 90$$ ways. The factors of $2!$ in the denominator represent the number of ways we could permute two indistinguishable objects within an arrangement without producing an arrangement that is distinguishable from the given arrangement.