My question reads as follows
Find the number of passwords that use each of the digits $3, 4, 5, 6, 7, 8, 9$ exactly once
In how many of the passwords
(a) are the first three digits even?
(b) are the three even digits consecutive?
(c) are the four odd digits consecutive?
(d) are no two odd digits consecutive?
Now for the first part I got $7!$ And then moved on to a. I did $3!4!$ and got $144$ but I am not too sure this is correct.
For b through d I'm confused as to what is meant by consecutive.
When I attempted b I got something like $7!-5(4!)$ because I took the total subtracted my five cases where we can have $468$ and the multiplied by $4!$ Since the odds still need to be arranged. I am not sure if my technique is correct so this is confusing me for c and d.
Your answers for the number of passwords and part a are correct. Now you need to allow your three consecutive even numbers to take any place in the password, not just the first three places, as long as they remain consecutive. So you still have $3!*4!$, but now you take into account the number of places the even numbers can take. What you have right now ($7!-5*4!)$ counts the number of passwords that don't have $468$ occurring in any place. Part c is similar to part b, but with the odd numbers. For part d, note that you have more odd numbers than even numbers (exactly one more, to be precise). How can you arrange them to get a password where no two odd numbers are adjacent?