Pascal's triangle in Elixir (using recursion where possible).

triangle.exs

# Author: MrSaints
# License: MIT

defmodule Factorial do
    def recursive(n, acc) when n === 1 or n === 0 do
        acc
    end

    def recursive(n, acc) do
        temp = n - 1
        recursive(temp, acc * temp)
    end

    def recursive(n) do
        acc = n
        recursive(n, acc)
    end

    def enum(n) do
        Enum.reduce n..1, fn(x, acc) ->
            x * acc
        end
    end
end

defmodule PascalsTriangle do
    def element(row, column) when column === 0 or row === column do
        1
    end

    def element(row, column) when column === 1 do
        row + 1
    end

    def element(row, column) do
        result = Factorial.recursive(row) /
        (Factorial.recursive(column) * Factorial.recursive(row - column))
        trunc result
    end

    def row(n, current, acc) when current > n do
        acc
    end

    def row(n, current, acc) do
        newTuple = Tuple.insert_at(acc, current, element(n, current))
        row n, current + 1, newTuple
    end

    def row(n) do
        row n, 1, {element(n, 0)}
    end

    def rows(n, current, acc) when n === current do
        acc
    end

    def rows(n, current, acc) do
        rows n, current + 1, acc ++ [row(current)]
    end

    def rows(n) do
        rows n, 1, [row(0)]
    end
end

You could have written:
def pascal(n), do: pascal(n, [1, 1])
defp pascal(0, _), do: [1] # Not really coherent -- Should be 1 but it is easier for later use
defp pascal(1, acc), do: acc
defp pascal(n, acc) do
r = Enum.map(Enum.zip(acc ++ [0], [0 | acc]), fn {x, y} -> x+y end)
pascal(n-1, r)
end

Here is another way:

defmodule PascalTriangle do
  def draw(1), do: (IO.puts("1"); [1])
  def draw(current_level) do
    list = draw(current_level - 1)
    new_list = [1] ++ for(x <- 0..length(list)-1, do: Enum.at(list, x) + Enum.at(list, x+1, 0))
    Enum.join(new_list, " ") |> IO.puts
    new_list
  end
end

PascalTriangle.draw(5)