/*
Description:
In the drawing below we have a part of the Pascal's triangle, lines are numbered from zero (top). The left diagonal in pale blue with only numbers equal to 1 is diagonal zero, then in dark green (1, 2, 3, 4, 5, 6, 7) is diagonal 1, then in pale green (1, 3, 6, 10, 15, 21) is diagonal 2 and so on.

We want to calculate the sum of the binomial coefficients on a given diagonal. The sum on diagonal 0 is 8 (we'll write it S(7, 0), 7 is the number of the line where we start, 0 is the number of the diagonal). In the same way S(7, 1) is 28, S(7, 2) is 56.

Can you write a program which calculate S(n, p) where n is the line where we start and p is the number of the diagonal?

The function will take n and p (with: n >= p >= 0) as parameters and will return the sum.

##Examples:

diagonal(20, 3) => 5985
diagonal(20, 4) => 20349
##Hint: When following a diagonal from top to bottom have a look at the numbers on the diagonal at its right.

##Ref: http://mathworld.wolfram.com/BinomialCoefficient.html
*/
function diagonal(n, p) {
  let fac = n => (n === 0) ? 1 : n * fac(n - 1);
  return Math.round(fac(n + 1) / (fac(n - p) * fac(p + 1)));
}
