For example:
There are two green marbles and two red marbles in a bag. Two marbles are chosen at random, what is the probability that the two marbles which have been chosen are of the same colour?
Ordered + distinct:
Marbles = $G1, G2, R1, R2$
Event space = $\{|G1~G2|, |G2~G1|, |R1~R2|, |R2~R1|\} = 4$ ways
Sample space = $^4P_2 = 12$ ways
Probability = $\dfrac{4}{12} = \dfrac{1}{3}$
Unordered + distinct:
Marbles = $G1, G2, R1, R2$
Event space = $\{|G1~G2|, |R1~R2|\} = 2$ ways
Sample space = $^4C_2 = 6$ ways
Probability = $\dfrac{2}{6} = \dfrac{1}{3}$
Ordered + identical:
Marbles = $G, G, R, R$
Event space = $\{|G~G|, |R~R|\} = 2$ ways
Sample space = $\{|G~G|, |G~R|, |R~G|, |R~R|\} = 4$ ways
Probability = $\dfrac{2}{4} = \dfrac{1}{2}$
Unordered + identical:
Marbles = $G, G, R, R$
Event space = $\{|G~G|, |R~R|\} = 2$ ways
Sample space = $\{||G~G|, |G~R|, |R~R||\} = 3$ ways
Probability = $\dfrac{2}{3}$
Which approach is the correct one? And more importantly, why are the other approaches wrong?
All of the methods can be done to find the probability.
Your problem in the "identical" approaches is that you are considering each choice as equally likely, whereas this is not the case. Using the unordered identical example, picking $GG$ has a probably of $\frac16$, but the probability of picking $GR$ is not (it's $\frac23$) There's a higher chance of picking one of each ball rather than picking two of the same color.
So you would want to add the probabilities of $GG$ and $RR$ which will come out to the correct answer of $1/3$.
It can be done similarly for the ordered identical approach.