/sci/ - Science & Math

Hi, I'd like some help with b), please

My method to solving it was this:
When A finishes 1st there are, 7! ways of arranging them so that J finishes after A,
If A finishes in 2nd place there are, 6! ways J can finish behind A. Don't multiply by two since one of them is included when A finishes first
A=3, 2!*5!,
A=4, 3!*4! and so on result: 7!+6!+2!5!+3!4!+4!3!+5!2!+6!
My answer is wrong, and I don't really understand where in my thought process I went wrong.