What is 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 is the probability that at least one person doesn’t get any object equal to $\frac{7^6-6!}{7^6}$?
Proof: A particular object has $6$ choices, i.e. $6$ people to select $1$ from. Similarly, every object has $6$ choices, so total ways are $6^6$.
You might have selected a person and calculated choices for him: $6$ for $6$ objects and $1$ when he does not get any object. This gave you $7^6$, which is incorrect.
So, total ways (without restriction) are $6^6$.
And, number of ways when each person gets $1$ object is $6!$, which you got.
So, required number of ways are $$6^6-6!$$ giving you a probability of $$\frac{6^6-6!}{6^6}$$