The total number of ways which 5 balls of different colors can be distributed among 3 persons so that each person gets at least one ball is..?
My attempt:
So, first we can choose any three balls to allocate to any person, so $\binom{5}{3}$ and, then distribute them then in 3! ways. And, so we have, $\binom{5}{3} \cdot 3! $. Now I have two remaining balls and this I can give randomly to any of the three people. For the first ball I have three choices and so do I for the second ball.
Hence, The net number of cases is
$ \binom{5}{3} \cdot 3! \cdot 3^2$
Now, I know that like the thing can happen in reverse. For example distributing three then distributing two, I could have some overlapping cases where the items which are distributed in the distributing two case is actually distributed in the distributing three step.
How do I account for the overlapping case?
Ans: 150
Sticks and stones attempt:
We have five objects and introduce two objects as dividers, subtract cases where one guy gets nothing (3 x 6!) and add cases where two people get nothing and third guy everything (3)
What I am looking for : A fix to this method such that it gives me the correct answer, in the other stack post linked to this one, it has ample amount of methods based on other ways to solve it but my question here is how to fix this method.
Best Answer
From my experience, when dealing with distinct objects, it is never a good approach to say "let's first distribute some to fulfill the condition, then distribute the rest without limitations" exactly because the overcounting you mentioned.
The correct approach is the inclusion-exclusion principle: $$3^5-3\cdot 2^5+3=150$$ We consider all the distributions possible, removing the one in which only two people can get the object (there are 3 ways to choose these two) and add the ones in which only one person can get the object (there are 3 ways to choose this one).