In how many ways can $7$ different objects be divided among three persons so that either one or two of them do not get any objects? 
In how many ways can $7$ different objects be divided among three persons so that either one or two of them do not get any objects?

My Approach
I am not able to understand how to solve the problem.
I did $7$!/$3$! . $4$!

Can anyone guide me how to solve the problem?


Call the people A, B,  and C. Suppose C is to be shut out. Then A can get any of the $2^7$ subsets of the set of objects.
Our first estimate of the number of ways is therefore $3\cdot 2^7$. However, this sum counts $6$ times the number of ways two of the people receive nothing.  However, there are only $3$ such ways, so the total is actually $3\cdot 2^7 -3$.
Another way: There are $2^7-2$ ways that C is shut out of the game and A and B  each receive at least one object. Multiply by $3$, and add the three ways in which one of the players gets everything.
Remark: Our first way used the principle of Inclusion/Exclusion, because it is such a useful idea. The second way, for this problem, is more straightforward. 
