I'm struggling to understand this question, I have determined the probability of winning the jackpot by doing 1/(49 C 6) which is 1/13983816. But I'm not too sure what the question is asking of, wouldn't the total number of available winning sequences be 1 if there's only one winning ticket. I'm not sure how to begin the question. Any help will be appreciated! Thanks!
Question: In the lottery game Lotto 6/49 a set of six numbered balls is randomly selected from a set of forty-nine. A person wins the grand prize if the six numbers on their ticket match the six numbers of the randomly selected balls. Adam and Michael are both avid players and decide to compare the total number of available winning sequences to the number of ways a subset of any 6 numbers can be selected. should the answers for both calculations be different?
Edit: Since the order does not matter, 6 numbers could be selected and they can be placed in any order thus 6! meaning there's 720 number of winning available sequences. How could I find the number of ways a subset of any 6 numbers can be selected?
The idea of asking this question is quite obscure.
As you have said, the probability of winning in such a scenario $ = \dfrac1{\binom{49}6}$
In the second part, you are asked to compare the number of winning sequences to the number of ways a subset of any $6$ numbers can be selected.
Well, the number of winning sequences will be $6!$, and the number of possible subsets of any $6$ numbers selected are $\binom{49}6$
What purpose such a comparison will serve is not clear !