arranging 2 blue balls, 2 red balls and 1 green ball How many ways are there to arrange 2 blue balls, 2 red balls and 1 green ball?

My Answer: $$\frac{5!}{2!*2!}$$

If this is incorrect then please help me understand where I am wrong.
But if this is correct then please help me with this:

"A bag holds $4$ red marbles, $5$ blue marbles, and $2$ green marbles. If $5$ marbles are selected one after another without replacement, what is the probability of drawing $2$ red marbles, $2$ blue marbles, and $1$ green marble?"

Correct Answer:
$$\binom{5}{2}\binom{3}{2}\binom{1}{1}\left(\frac{4}{11}\right)\left(\frac{3}{10}\right)\left(\frac{5}{9}\right)\left(\frac{4}{8}\right)\left(\frac{2}{7}\right) = \frac{20}{77}$$

why does it have this?
$$\binom{5}{2}\binom{3}{2}\binom{1}{1}$$
instead of this:
$$\frac{5!}{2!*2!}$$

for more info, you can refer to my pervious question here:
Basic combinations logic doubt in probability
It is because they are in fact equal.$$\begin{align}\binom 5 2\binom 32\binom 11&amp;=\dfrac{5!}{2!~3!}\dfrac{3!}{2!~1!}\dfrac{1!}{1!~0!}\\[1ex]&amp;=\dfrac{5!}{2!~2!~1!}\end{align}$$

Some just prefer presenting such expressions as binomial coefficients as it reflects the logic used to derive the expression.

I rather the multinomial form. $$\binom 5{2,2,1}$$Use whatever you find most comfortable.
This is called The Mississippi Formula and yes you are correct! Basically you are getting rid of the "overcounts" by dividing by 2! 2!