As the title asks, how many ways are there to permute the elements of the set $[7]$ so that an even number is in the first position?
This is a question from "A Walk Through Combinatorics" by Bona.
Update: I think I may have figured it out, but there are no answers to this in the book, so some verification would be great. I know the total number of possible permutations of $[7]$ is $7!$. And it seems to me that the chances of an even number being in the first position would be $\frac{3}{7}$, so the total number of permutations with an even number in the first position would be $\frac{3}{7}(7!) = 2160$. Is this correct? If so, it was much simpler than I thought.
I think the answer should be 3(6!). Because there are only 3 ways to chose the first number then the second is 6 then 5 and so on.