I had some doubt regarding below problem.

The number of ways of distributing 6 objects to 6 persons such that at least one of them does not get anything.

I worked like

Without restriction total ways= $ 7^6$

Number of cases in which all persons get something=$ 6!$

Number of cases where at least one person does not get anything=$ 7^6-6!$

Am I correct?

Also if the probability is asked for at least one person don’t get any object then should it be equal to $ \frac{7^6-6!}{7^6}$ ?