A certain state's lottery consists of choosing $6$ numbers at random without replacement from the set $\{1,2,3,\ldots,40\}$. What is the probability that someone wins the jackpot?
I just want to make sure my answer of $\left( \frac{1}{40} \right)^6$ is the correct approach to the situation. Whatever that answer is I have not computed it yet.
Assuming order is not important, there are $40\choose 6$ ways to choose the six numbers. So the probability of getting the correct six numbers with one ticket is $\frac{1}{40\choose 6}=\frac{(6!)(34!)}{40!}$