10 objects are to be distributed among 10 people randomly. What is the probability that at least one of them will get nothing?
Now, I had thought of two ways of doing it:
I assumed here objects to be identical. Hence total number of ways in which 10 objects would be distributed among 10 people is $\frac{19!}{9!10!}$. Further, the total number of ways equals every one gets something + at least one gets nothing. And, there is only one way in which every one gets something. Hence the number of ways in which at least one person gets nothing is $\frac{19!}{9!10!} - 1$. The required probability thus equals: $\frac{\frac{19!}{9!10!} - 1}{\frac{19!}{9!10!}}$.
I assumed all objects to be distinct. Hence number of total ways is $10^{10}$. Also, total ways = every one gets something + at least one gets nothing. Further, every one gets something in 10! Ways. Hence, number of ways in which at least one gets nothing = 10^10-10!. Hence the required probability is: $\frac{10^{10}-10!}{10^{10}}$.
The problem is that my answers are different for these two methods.
I strongly feel that the answer should be same irrespective of whether they are distinct or identical. How and where am I going wrong? Why is the probability of quantity dependent on the quality of the objects?
The probability of at least one of the people getting nothing, equals 1 minus the probability of all of them getting something. There are $10^{10}$ equally likely ways to distribute the objects, and only $10!$ valid ways in which this can be done. We indeed find:
$$1 - \frac{10!}{10^{10}}$$
The problem with your other approach, is that the probabilities of the ${19 \choose 9}$ possible combinations are not the same. For instance, the probability of all gifts being given to the first person equals $\frac{1}{10^{10}}$, while the probability of each person getting something equals $\frac{10!}{10^{10}}$.
To make this more tangible, try doing the same for two people and two objects. The probability of the first person getting both items equals $\left(\frac{1}{2}\right)^2 = \frac{1}{4}$, while the probability of both people getting one object equals $1 \cdot \frac{1}{2} = \frac{1}{2}$. Indeed, the probabilities of the three possible combinations are not the same.