Lottery game exercise probability I am not sure whether my answer to this problem is correct or not.
Suppose that 100 people enter a contest and that different winners are selected at random for first, second, and third prizes. What is the probability that Kumar, Janice, and Pedro each win a prize if each has entered the contest?
The sample space has to be $|S|= $$100\choose3$ as it is number of all possible winners.
I think the event should be $|E|= $$3\choose3$, but then we are ignoring the who's winning what so is it $P(3,3)$?
 A: your answer, $\dbinom{3}{3}$ for the event and $\dbinom{100}{3}$  for the sample space is correct.
If you are concerned about the order, say, $P(3,3)$ for the event, then the order must also applying the sample space, which is then $P(100,3)$ which is same the above case.
A: We have that there are $\binom{100}{3}$ different choices of $3$ people to be the winners, and for each such choice, there are $3!$ in which their orders can be chosen. Hence, $|S|=\binom{100}3\cdot3!$.
On the other hand, there are $3!$ success cases for our problem, corresponding to the number of ways we can order the winners Kumar, Janice and Pedro. Hence, $|E|=3!$ and the probability is
$$\frac{|E|}{|S|}=\frac{97!\cdot3!}{100!}=\frac{3!}{100\cdot99\cdot98}=\frac{1}{50\cdot33\cdot98}\simeq0.000618\%$$

We see that the order ends up not mattering (the $3!$ cancels out). That's because only care about Kumar, Janice and Pedro getting a prize each. In other words, we are not particularly interested in an order for our winners.
A: 100*99*98/6=161700
answer is one out of  161700
