How many ways are there to go through all intersections and come back to starting point in the following street map... 
The street map of a district in a city is as follows(3 horizontal and 10 vertical streets):
  

A policeman wants to visit each intersection starting from the upper right corner and get back to his starting point provided that each intersection is visited just once.In how many ways is this possible??
The condition given confused me...
 A: Consider the 9 sections of horizontal roads along the middle of the grid.
Suppose the policeman travelled along two consecutive sections. To visit the intersection on the top and the intersection on the bottom street he must also go horizontally along both of those streets. It is then not possible to get back to his starting point. Therefore he does not go along two consecutive middle sections.
Suppose there are two consecutive middle sections that the policeman did not travel along. To visit the intersection in the middle he would have to go vertically from the top to the bottom street (or vice versa). This also makes it impossible for him to get back to his starting point. Therefore there are not two consecutive middle sections that he does not go along.
This means that the middle road sections are alternately travelled and untravelled. The first and last middle sections cannot be travelled because then he cannot visit both corners on that end of the grid.
So there are five middle road sections that he does not travel down. For each of those he will have to travel along both the top and the bottom horizontal sections of road. There are also four middle road sections that he does travel down. For each of those he will also have to travel either along the top horizontal road or the bottom one. This gives 16 possibilities.
All the vertical movements are completely determined by which horizontal sections he goes along. So there are 16 possible circuits. If the direction of travel matters, there are 2*16=32 possible routes he can take.
Here's a picture showing the steps:

A: Lemma 1: When the policeman reaches the bottom left corner, he has at most one possible way back to his starting point.
Lemma 2: The path from starting point to the opposite corner separates the city in two parts. When the policeman reaches the bottom left corner, he has a way back only if he either visited all the South-East intersections and none of the the North-West intersections, or the other way around.
Let's suppose he has visited all the South-East intersections (we will find the other case by considering each parcours in the reverse direction).
He has necessarily visited A1, B1, ..., I1, J1 and J2. He may have visited both H2 and I2 or none of them, but not one of them only. The same applies for the couples (B2,C2), (D2,E2) and (F2,G2).
His path is uniquely determined by which of these four pairs of intersections he has, or not, visited.
This means $4$ binary choices, so the total number of such paths is $2^4=16$.
Since he can also parcour each of these path in the other way (visiting the NW corner before the SE), the total number of possible paths is $2*16=32$.  
A: If you look at a problem and have no idea how to begin, try solving a simpler problem.
You might consider fewer horizontal streets. Clearly there must be at least two, otherwise it is impossible to return to the start.
Now consider a grid with two horizontal streets and ten vertical ones.
Try drawing a path from the upper right corner along the edges of the grid and back to the upper right corner without touching an intersection more than once.
Did you touch every intersection? How can you touch every intersection, but only once?
(I find it hard to see the path I have drawn on a grid because the lines I drew as part of the "path" tend to blend into the lines that were part of the original grid. It seems easier to just draw dots on the paper representing all the intersections of the grid, and then draw a path connecting the dots using only horizontal or vertical lines.)
You should find that for a grid with just two horizontal streets, you have very few choices about how to draw the path without missing any of the intersections or visiting any of them a second time.
Now try three horizontal streets and three vertical ones.
You should quickly find it cannot be done. Why not?
Now three horizontal streets and four vertical ones. How many ways?
You can find out by drawing each path.
What about three horizontal streets and five vertical streets?
What about three horizontal streets and six vertical streets?
Eventually you may see a pattern that you can generalize to three horizontal streets and as many vertical streets as you like, and apply this pattern to the original problem, three horizontal streets and ten vertical ones.
