Say I have $5$ distinct objects which I need to place in $3$ identical bins such that 1st bin contains $3$ objects, 2nd $2$ objects and 3rd none. I get confused between the cases of $3$ "identical bins" case and $3$ distinct bins case. I don't know whether my approach is correct or not. Please have a look and correct me if i am thinking wrongly.
In case of $3$ distinct bins, I label bins as 1,2,3 (in case of identical bins labelling not allowed). Number of ways $= 5C3 \cdot 2C2 \cdot 3!$.
In case of $3$ identical bins, labelling of bins is not allowed. It means we do not know which bin we picked. Now number of ways here $= 5C3$ (select first $3$ objects out of $5$) $\cdot \ldots \to$ now how to place, I don't know where to place in what bin. If indistinguishable, how can I choose a bin and ensure that 1 bin contains $3$, another $2$ and other none?
Your answer for the number of ways of placing five distinct objects in three distinct bins is correct. Another way to see this is to choose which bin will receive three objects, choose which three of the five objects will be placed in that bin, and choose which of the other bins will receive the other two objects. $$\binom{3}{1}\binom{5}{3}\binom{2}{1}\binom{2}{2} = \binom{5}{3}\binom{2}{2}3!$$
If the bins are indistinguishable, choose which three of the five objects go in the same bin. Place the other two objects in another bin. It does not matter which of the other bins you choose since they are indistinguishable. Hence, there are $$\binom{5}{3}\binom{2}{2}$$ ways to distribute five distinct objects to three indistinguishable if three of the objects are placed in one bin and two of the objects are placed in another.