Answer :
Answer:
There are 3,921,225 ways to select the winners.
Step-by-step explanation:
This problem is about combinations with no repetitions, because the same person can't win four times. It's a combinaction because the order of winning doesn't really matter.
Combinations without repetitions are defined as
[tex]C_{n}^{r} =\frac{n!}{r!(n-r)!}[/tex]
Where [tex]n=100[/tex] and [tex]r=4[/tex].
Replacing values, we have
[tex]C_{100}^{4} =\frac{100!}{4!(100-4)!}=\frac{100!}{4! 96!}=\frac{100 \times 99 \times 98 \times 97 \times 96!}{4! \times 96!}= \frac{94,109,400}{24}= 3,921,225[/tex]
Therefore, there are 3,921,225 ways to select the winners.
Additionally, as you can imagine, the probability of winning is extremely low, it would be 3,921,225 to 1.