First of all here balls are indistinguishable. Let us consider the scenario where we have two indistinguishabe balls and number of distinguishable bags are 3 i.e. m=2 and n=3.
Note: Since number of balls are less than number of bags, i have assumed that no bag can contain more than one ball. There can be another case where a bag can contain more than one ball.
But if number of balls are greater than number of bags, then definitely a bag must contain more than one ball.
Thus there are 3 ways in which 2 bags can be selected out of 3 bags. They are (bag1,bag2) , (bag1,bag3), (bag2,bag3).
Let us consider the case of (bag1, bag2). There is only one way in which two balls can be placed in these bags. Why?
There are two possibilities:
1. Placing 1st ball in bag1 and 2nd ball in bag2.
2. Placing 1st ball in bag2 and 2nd ball in bag1.
Both these possibilities are same since balls are indistinguishable. (B,B) and (B,B) are same where B denotes ball.
Number of possibilities for placing m indistinguishable balls in n distinguishable bags considering no bag can contain more than one ball= nCm