Given the question:
"How many ways we can distribute $10$ distinct objects among $4$ persons such that each person gets at most $9$ objects?"
The correct answer given is $4^{10} -4$. We have subtracted the 4 cases in which each person gets all 10 objects.
Can't the answer be $10 \cdot 4^9$? We have ${10 \choose 1}$ ways of selecting an object. We remove that object and now each object has $9$ choices. So each object can now get at most $9$ objects only.
First, I propose a different derivation of the answer $4^{10}-4$ that goes into the direction of your way of thinking.
Let us say that we distribute the first 9 objects.
There are $4^9$ ways to do so. In all those cases except 4, i.e. $4^9-4$, each person received at most 8 objects, so we can safely distribute the last one as we prefer, so in 4 different ways.
We get a total of $(4^9-4)\cdot4$ total ways of distributing the objects.
We are left with the 4 cases in which someone already received 9 objects.
In these 4 cases, we can distribute the remaining object to a different person, so in 3 different ways.
The total is thus: $ (4^9-4)\cdot 4 + 4\cdot 3 , $ which is equal to $4^{10}-4$.
Then, your proposal is to start by removing one of the 10 objects, not necessarily the last one, say the $n$th. You have 10 different ways of doing so.
Then, for each $n$, we follow the reasoning above and we get $4^{10}-4$ different ways. But you get all the different ways of distributing for each $n$, so you do not have to multiply times 10.
