Probability of visiting $4$ cities 
On her vacations Veena visits four cities $(A, B, C\ \text{and}\ D)$ in a random
  order. 
What is the probability that she visits
(i) $A$ before $B$? 
(ii) $A$ before $B$ and $B$ before $C$?
(iii) $A$ first and $B$ last? 
(iv) $A$ either first or second?
(v) $A$ just before $B$?

The number of ways Veena can visit four cities is $4!=24$
$S = \{ABCD, ABDC, ACBD, ACDB, ADBC, ADCB\\
BACD, BADC, BDAC, BDCA, BCAD, BCDA\\
CABD, CADB, CBDA, CBAD, CDAB, CDBA\\
DABC, DACB, DBCA, DBAC, DCAB, DCBA\}$
I want to know if their is an easier way than writing all these $24$ possibilities .
I know their might be some permutation and combination way to solve it easily but I can't figure it out.
I have studied maths upto $12$th grade.
 A: Let's start with part $(i)$ (the other parts will use the same idea):
If Veena is to visit $A$ before she visits $B$, then we are looking to count the number of strings using $A,B,C,D$, where $A$ precedes $B$. Such strings could have the form
$$AB__$$
$$A_B_$$
$$A__B$$
$$_AB_$$
$$_A_B$$
$$__AB$$
We didn't actually need to list these; we could have simply recognized that there are 3 possible positions for $A$ (any but the last), and that if $A$ is in position $i$ ($i = 1,2,3$), then there are $4-i$ possible positions for $B$. Therefore the number of such strings of this general form is $\sum_{i=1}^3(4-i) = 3+2+1 = 6$, which agrees with the list above.
Now all we have to do is place $C$ and $D$, and there are no restrictions on how we may do this. There are two ways to place $C$ and $D$ in the empty spaces in each string listed above, therefore we get a total of $2\cdot6 = 12$ unique strings where $A$ precedes $B$.
Thus the probability that Veena visits $A$ before she visits $B$ is
$$\frac{\text{# of strings where $A$ precedes $B$}}{\text{total number of strings}} = \frac{12}{4!} = \frac{1}{2}$$
For parts $(ii)$ - $(v)$, the same general method is employed, we are just putting different restrictions on the particular types of strings we are looking for.
A: For the first one, which is more likely, A before B or B before A? Keep in mind one of these two has to occur. So P(A before B)+P(B before A)=1. Would it make sense for one of these to have a higher probability than the other?
For the second one, we know she visits and and then B and then C. The only question is, where does D fit in? How many places can you put D in between or outside ABC?visualize this as _A_B_C_. The D has to go into one of those open slots. How many slots are there?
For the third, if A is first and B is last, how many ways can we fill in the middle? We can visualize this as A_ _B. How many ways can I put C and D in the two open slots?
For the fourth, we can use similar reasoning to the first. Is it more likely for A to be (first or second) or (third or fourth)? But one of the two has to happen. What does that tell you about the probability?
For the last one, if A is just before B, then we can think of AB as a single object. Then it's how many ways can we visit three cities?(well... Two cities and one double city) then divide by 24. This one is $\frac{3!}{4!}=\frac{1}4$ as you stated in a comment.
Let me know if you still have questions.
A: 
(i) A before B?

$50\%$ she'll visit $A$ before $B$ and $50\%$ she'll visit $A$ after $B$, so the probability is $\frac{1}{2}$.


(ii) A before B and B before C?

Arrange $ABC$ by that order, and then stick $D$ anywhere. There are $4$ ways to do it:


*

*$ABCD$

*$ABDC$

*$ADBC$

*$DABC$


Hence the probability is $\frac{4}{24}=\frac{1}{6}$.


(iii) A first and B last?

Arrange $A$ first and $B$ last, and then stick $C$ and $D$ between them. There are $2$ ways to do it:


*

*$ACDB$

*$ADCB$


Hence the probability is $\frac{2}{24}=\frac{1}{12}$.


(iv) A either first or second?

$A$ can be located in either one of $4$ places, with equal probability for each one of them.
Hence the probability is $\frac{2}{4}=\frac{1}{2}$.


(v) A just before B?

This description is rather tricky.
One might interpret it as if $A$ has to come before $B$.
I believe that the actual meaning is that $A$ is allowed to come only before $B$.
There are $2$ combinations in which $A$ comes only before $B$:


*

*$CDAB$

*$DCAB$


And there are $6$ more combinations in which $A$ doesn't come before any letter:


*

*$BCDA$

*$BDCA$

*$CBDA$

*$CDBA$

*$DBCA$

*$DCBA$


Hence the probability is $\frac{2+6}{24}=\frac{1}{3}$.
A: The probability that she visits A before B must be $1$ minus the probability that she visits B before A.
Suppose we find the probability that she visits A before B.
Then we rename the cities, so the one we called A we now call B and vice-versa. What is then the probability that she visits A before B?  It's the same math problem, so the probability must be the same number.  But now it's the probability that she visits the city we initially called B before she visits the city we initially called A.
Therefore the probability that she visits A before B must be the same as the probability that she visits B before A, and the sum of those two (equal) numbers must be $1$.  Hence that probability is $1/2$.
That is an argument from symmetry.
