## Question

Starting in the top left corner of a 2x2 grid, there are 6 routes (without backtracking) to the bottom right corner.

How many routes are there through a 20x20 grid?

## Commentary

The grid can be expressed as Pascal’s Triangle:

```
1
1 1
1 (2) 1
1 3 3 1
1 4 (6) 4 1
1 5 10 10 5 1
1 6 15 (20) 15 6 1
```

Note that the solution for a 1x1 grid is 2, a 2x2 grid is 6, and a 3x3 grid is 20.

If we compare these solutions to Pascal’s Triangle, we see that they correspond to the 1st element in the 2nd row, the 2nd element in the 4th row, and the 3rd element in the 6th row, respectively. (Note that Pascal’s Triangle is zero-indexed.)

The binomial coefficient $\binom {n} {k}$ can be used to determine the $k$th element in the $n$th row of Pascal’s Triangle. Thus, we could express the aforementioned solutions as $\binom {2} {1}$, $\binom {4} {2}$, and $\binom {6} {3}$, respectively.

Thus, a general solution for grids of size $x$ is

$routes = \binom {2x} {x}$.

## Haskell

```
factorial :: Integer -> Integer
factorial n = product [1..n]
choose :: Integer -> Integer -> Integer
choose n k = div (factorial n) $ factorial k * factorial (n - k)
main :: IO ()
main = print $ choose 40 20
```

## Python

## Ruby

```
#!/usr/bin/env ruby
class Integer
def choose(k)
(self-k+1 .. self).inject(1, &:*) / (2 .. k).inject(1, &:*)
end
end
puts 40.choose(20)
```