Railway Station

Knowing that you have to make exactly 8 moves, you will never end at B or D.
Half of the time you will end up at A and the other half of the time you will end at C.
Since we have 2 directions to go and 8 moves to make we have 2^8 = 256 total paths.
We said earlier we expect half of those end us back at A, so our answer should be 128.

The adjacency matrix of the graph is

$M = \begin{pmatrix} 0 & 1 & 0 & 1\\ 1 & 0 & 1 & 0\\ 0 & 1 & 0 & 1\\ 1 & 0 & 1 & 0 \end{pmatrix},$

if the rows and columns are ordered $A, B, C, D$ . Each entry in $M^n$ is the number of paths of length exactly $n$ between the vertices corresponding to its row and column. By direct computation,

$M^8 = \begin{pmatrix} 128 & 0 & 128 & 0\\ 0 & 128 & 0 & 128\\ 128 & 0 & 128 & 0\\ 0 & 128 & 0 & 128\\ \end{pmatrix},$

so the number of paths of length 8 from $A$ to $A$ is the top left entry, 128.

$\begin{aligned} \text{Let us}&\text{ define a function } f(n) \text{ such that,}\\ f(n)&={\begin{cases}\hspace{3mm} 1&:\text{if the nth ticket was used to move in the clockwise direction}\\ -1&:\text{if the nth ticket was used to move in the anti-clockwise direction}\\ \end{cases}}\\ \text{Now you}&\text{ can start and end the journey at A if ,}\\ \sum_{n=1}^8 f(n)&\equiv0\pmod{4} \hspace{4mm}\color{#3D99F6}\small(1)\\\\ \text{Let }`\text{a'} &\text{ denote the number of }1\text{'s and }`\text{b'} \text{ denote the number of }-1\text{'s.}\\ a+b&=8 \hspace{4mm}\color{#3D99F6}\small\text{The total number of tickets}\\ a-b&\equiv0\pmod {4} \hspace{4mm}\color{#3D99F6}\small\text{From} (1),-8\leq(a-b)\leq8\\ \implies{a-b}&=0,4,-4,8,-8 \\ \implies(a,b)&=(4,4),(6,2),(2,6),(8,0),(0,8)\\\\ \text{The}&\text{ number of ways, of permuting } `\text{a' 1's and } `\text{b' -1's over 8 spaces is }\dfrac{8!}{a!b!}\\ \text{T}_{\small\text{ways}}&= \sum_{(a,b)} \dfrac{8!}{a!b!}\\ &=\dfrac{8!}{4! 4!}+\dfrac{8!}{6!2!}++\dfrac{8!}{2!6!}+\dfrac{8!}{8!0!}+\dfrac{8!}{0!8!}\\ &=70+28+28+1+1\\ &=\color{#EC7300}\boxed{\color{#333333}128} \end{aligned}$

A > [B/D] > [A/C] > [B/D] > [A/C] > [B/D] > [A/C] > [B/D] > A

     2     *     2     *    2     *    2     *    2     *    2     *    2     *  1  = 2^7  = 128

There are 2 choices in each of 7 middle cases, for going from A to A by using 8 tickets, and in last case, one must have to return in A. So, there is one choice, and hence the problem is done.