Seongcheol has $3$ red shirts and $2$ green shirts, such that he cannot tell the difference between his three red shirts and he similarly cannot tell the difference between his two green shirts. In how many ways can he hang them in a row in his closet, given that he does not want the two green shirts next to each other?
1. **Calculate the total number of ways to arrange the shirts without any restrictions:**

   Seongcheol has 3 red shirts (R) and 2 green shirts (G). The total number of shirts is 5. Since the red shirts are indistinguishable among themselves and the green shirts are indistinguishable among themselves, the number of ways to arrange these shirts is given by the multinomial coefficient:
   \[
   \frac{5!}{3!2!} = \frac{120}{6 \cdot 2} = 10
   \]

2. **Calculate the number of ways to arrange the shirts with the two green shirts next to each other:**

   If the two green shirts are next to each other, we can treat them as a single "block" or "big green shirt". This reduces the problem to arranging 4 items: 3 red shirts and 1 "big green shirt". The number of ways to arrange these 4 items is:
   \[
   \frac{4!}{3!1!} = \frac{24}{6} = 4
   \]

3. **Subtract the restricted arrangements from the total arrangements:**

   The number of ways to arrange the shirts such that the two green shirts are not next to each other is the total number of arrangements minus the number of restricted arrangements:
   \[
   10 - 4 = 6
   \]

Conclusion:
\[
\boxed{6}
\]