arranging 2 blue balls, 2 red balls and 1 green ball

Question

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

Answer 1

Well, $$ \color{blue}{\binom 52 \binom 32 \binom 11 = \frac{5!}{2! \times 2!} = 30} $$

So it seems that what you are doing is also correct. However, your thought processes when computing the answer were different, so I will just highlight that.

• What was the person who wrote the "correct answer" thinking? He was thinking : let me first decide when the blue balls were drawn, followed by when the red balls were drawn, followed by when the green ball was drawn. So what he did was this : the first $\binom 52$ represents the two chosen spots in which the red balls were drawn. Then, these spots are gone, so from the remaining three spots, two were chosen for the red balls, and then the one remaining spot for the green ball. This is why the answer is written in that order as well.

• What were you thinking? Probably : you assumed all the balls are different, then the number of orders in which they can appear is $5!$. Now, you remembered that two of them are blue and two of them are red, so for this you divided by $2!$ twice to account for that.

And your answers are the same, because both the ways of thinking about the problem are correct, and therefore will lead to the same answer!

Answer 2

It is because they are in fact equal.$$\begin{align}\binom 5 2\binom 32\binom 11&=\dfrac{5!}{2!~3!}\dfrac{3!}{2!~1!}\dfrac{1!}{1!~0!}\\[1ex]&=\dfrac{5!}{2!~2!~1!}\end{align}$$

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.

Answer 3

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