Arrangement probabilities I worked out each part, but I am not really sure about my answers after part a.
How many different linear arrangements are there of the letters A B C D E F for which:
a. A and B are next to each other? 2!5! = 240
I grouped A and B as "one element" so now there's 5 elements, so 5! ways to arrange them, multiplied by the 2! ways to arrange AB (either AB or BA)
b. A is before B? 
correct answer: 6!/2
1st attempt:
5!+4!+3!+2!+1!= 153
The way I pictured this problem is:
A _ _ _ _ _ 5! ways for B to be arranged
X A _ _ _ _ 4! ways for B to be arranged
X X A _ _ _ 3! ways for B to be arranged
X X X A _ _ 2! ways for B to be arranged
X X X X A _ 1! ways for B to be arranged
c. A is before B and B is before C? 
1st attempt:
$5!4!(3!)^22!1!$
5! ways to choose letter A
4! ways to choose letter B
3! ways to choose letter C
3! ways to choose letter 4
2! ways to choose letter 5
1! way to choose letter 6
d. A is before B and C is before D? 
1st attempt:
(I am not at all confident about this answer)
5!4! ways for A to be before B, then 3!2! ways for C to be before D, multiplied by 2! because we could do the reverse (5!4! ways for C to be before D, then 3!2! ways for A to be before B
e. A and B are next to each other and C and D are also next to each other? = $4!2!2!$
So I am grouping AB as "one element" and CD as one element", so there are now 4 element. 
4! ways to arrange them 
2! ways to arrange AB
2! ways to arrange CD
f. E is not the last in line? = $(5!)^24!3!2!1!$
5! ways to choose E 
5! ways to choose letter 2
4! ways to choose letter 3
3! ways to choose letter 4
2! ways to choose letter 5
1! ways to choose letter 6
 A: a) Good.
b) There are many ways of counting. But it is easiest to note that of the $6!$ permutations, by symmetry half have A before B and half have A after.
c) For any of the  $3!$ possible orders $\sigma$ that A, B, C can come in, there are equal numbers of permutations of A, B, C, D, E, F in which A, B, C occur in the order $\sigma$. So for the count we divide $6!$ by $3!$.
d) For a change, let's argue probabilistically. The probability that A comes before B is $\frac{1}{2}$. Given that A comes before B, the probability C comes before D is $\frac{1}{2}$. There are $6!$ equally likely permutations. The probability of A before B and C before D is $\frac{1}{2^2}$, so the number of "favourables" is $\frac{6!}{2^2}$.
e) Good
f) Count the complement, the words that end in E. There are $5!$ of them, so $6!-5!$ don't end in E.
Alternately, by symmetry the proportion of permutations that don't end in E is $\frac{5}{6}$.
There are only $6!$ permutations of the $6$ letters. Your proposed answer is much larger than $6!$. 
Remark: Symmetry is your friend, exploit it. (I had better work on the wording, that didn't come out sounding quite right.)
