This is a Math problem for exercising. I solved it and went through it with my teacher. However, I still don't understand when I should use $3!$ and $3$ (originally $\frac{3!}{2!}$)
The problem (translated from French):
An urn contains 9 balls: three are red and numbered $-1, -1, 1$; two are green and numbered $-2, 2$ and four are white and numbered $1, -2, 2, 2$. All the balls are indiscernible to the touch.
We draw three balls successively and with replacement. Calculate the probability of
G : «Having a negative product of which only one of the three balls is numbered $-2$»
The solution:
There are two possibilities: either we draw (-2), (+), (+) or (-2), (-1), (-1)
$$P(G) = 3\left(\frac{2}{9}\times\frac{5}{9}\times\frac{5}{9}\right) + 3\left(\frac{2}{9}\times\frac{2}{9}\times\frac{2}{9}\right)$$
The question here is, should we use $3$ or $3!$? As I understand $3$ is used because $\frac{5}{9}$ and $\frac{5}{9}$ will give as the same result regardless of the color of the drawn ball. So, we multiply $\left(\frac{2}{9}\times\frac{5}{9}\times\frac{5}{9}\right)$ by $\frac{3!}{2!} = 3$.
In the next problem, we used $3!$, although it was generally the same problem.
An urn contains six red balls numbered $-1, -1, 0, 1, 1, 1$ and three yellow balls numbered $-1, 0, 1$. We draw successively and without replacement three balls. Calculate the probability of
G : «The sum of three numbers written on the three drawn balls is equal to $1$»
Solution: $0+0+1$ or $-1+1+1$
$$P(G) = 3!\left(\frac{2}{9}\times\frac{1}{8}\times\frac{4}{7}\right) + 3!\left(\frac{3}{9}\times\frac{4}{8}\times\frac{3}{7}\right)$$
What I didn't understand is why did we use $3!$? Shouldn't the color of the ball be neglected here too? If yes, then why didn't we use $3$?
Side note: I'm a high school student. "Majoring" in Mathematics. We don't use a lot of formulas in probability for this year. We just use the basic concepts and formulas as I've shown in the solutions above.
And, can someone please suggest a better name for this question?
Thank you very much.