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

This is called The Mississippi Formula and yes you are correct! Basically you are getting rid of the "overcounts" by dividing by 2! 2!
