Getting Round the City

Annette from Copleston Sixth Form explains why there are $56$ possible routes from $A$ to $B$:

All routes between $A$ and $B$ will be equal in length as you need to
travel $3$ east and $5$ north in any order. In total there are $56$ different
paths leading to $B$.

$1$	$6$	$21$	$56$ ($B$)
$1$	$5$	$15$	$35$
$1$	$4$	$10$	$20$
$1$	$3$	$6$	$10$
$1$	$2$	$3$	$4$
$A$	$1$	$1$	$1$

The number in each square represents the number of possible paths from $A$ to that square.

It is worked out like this:

$M$	$M+N$
-	$N$

 
because you must arrive at a square from the point either directly south or directly west of it.

Leah from St. Stephen's School Carramar found a general rule for the number of ways of travelling from point $(x_1,y_1)$  to $(x_2,y_2)$ going only north and east and discussed some ways to find the number of paths from $A$ to $B$:


If the route must be taken without crossing diagonally, it will take $(x_2-x_1) + (y_2-y_1)$ 
units to reach the opposite point.

E.g the points $(0,1)$ and $(3,6)$, a path will be $(6-0)+(3-1) = 8$ units long.

Number of paths from $(x_1,y_1)$ to $(x_2,y_2)$ = number of paths consisting of $(x_2-x_1)$ steps east and $(y_2-y_1)$ steps north, in some order

$ =$ ${(x_2-x_1)+(y_2-y_1)}\choose{(x_2-x_1)}$ $= \frac{[(x_2-x_1)+(y_2-y_1)]!}{(x_2-x_1)!(y_2-y_1)!}$

How many paths are there from $A$ to $B$?

$\frac{(5+3)!}{5!3!} = 56$.

Alternative methods :
- Drawing a tree diagram via labeling each node as a
letter, which gives $56$ options.

-Drawing out all the possible routes.

Hannah from Burntwood explains how many different paths we can take if we only want to travel $6$ blocks:

A good way to think about this is as a tree diagram.

At the starting point, you have $4$ options of where to go.

Then at each subsequent point you have a choice of $3$ paths (assuming you can't double back on yourself)

So, the formula would be $4 \times 3^{(n-1)}$, where $n$=the number of blocks you walk.

In this case, $n=6$, so there are $4x3^5= 972$ paths.

Thank you to everyone who submitted a solution to this problem!