Help with math problem involving permutations (people lining up)? Here's how the question goes.
There are 12 students who are lining up to enter class.
a) How many arrangements are possible?
b) How many arrangements are possible, if: 
1) Martha and Jonny are among the last four students. 2) Ellis is in between Carter and Ethan and they are together. 3) Ellis is in between Carter and Ethan but they aren't necessarily together. 4) There are exactly 3 students between Anna and Josh. 
Okay, so the first one is simple (12!).
Things get tricky for me in the second part. No clue how to do b1. b2 is fairly simple, I think, it should just be 9! + 2 by treating the group as one. Zero clue for b3, I'm not sure how to do it. For b4 there are 8 ways to arrange the group of 5, so then I add 2 possibilities for Anna and Josh being on different sides and then add 10! for the order of the rest of the students. I'm not sure if this is correct.
Can someone give me some hints on how to tackle the other ones?
 A: In most of the subproblems you can place the distinguished students first, then the rest – the numbers of ways for each part are multiplied together to get the final result.
For part 1, Martha and Jonny can be arranged among the last four students in $4×3=12$ ways. Then there are $10!$ ways to place the remaining students, making $12×10!$ ways overall.
Part 2's answer is not $9!+2$; rather, it's $2×10!$, the $10!$ obtained by treating the specified block on par with the other nine students.
Part 3 requires some sneakiness. Replace the labels "Ellis", "Carter" and "Ethan" with some generic "Student to be named later". These labels, now identical, can be distributed across the line in $\binom{12}3$ ways. For each of these possibilities, Ellis, Carter and Ethan can then be assigned to these "name later" labels in two ways, and there are $9!$ permutations of the remaining students. Thus there are $3\binom{12}39!$ arrangements in total.
For part 4, there are indeed $8$ position sets for Anna and Josh, then two ways to permute them in that set, then $10!$ arrangements of the other students, making $16×10!$ ways overall.
Where you show an attempt, you are using addition to combine independent events; you should be using multiplication instead.
