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
![alternative text](http://i.imgur.com/eUGaNvIm.jpg)
from functools import reduce 
def comb(n, p):
    a,b = n-p, p
    return reduce(lambda x, y : x*y, range(max(a,b)+1,n+1))//reduce(lambda x, y : x*y, range(1,min(a,b)+1))

def diagonal(n, p):
    return comb(n+1, p+1)
from functools import reduce 
def comb(n, p):
    a,b = n-p, p
    return reduce(lambda x, y : x*y, range(max(a,b)+1,n+1))//reduce(lambda x, y : x*y, range(1,min(a,b)+1))

def diagonal(n, p):
    return comb(n+1, p+1)