Skip to main content
### Number and algebra

### Geometry and measure

### Probability and statistics

### Working mathematically

### For younger learners

### Advanced mathematics

# Getting Round the City

Or search by topic

Age 11 to 16

- Problem
- Student Solutions
- Teachers' Resources

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 $B$.

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

It is worked out like this:

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!

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!