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.