Problem $1.[30$ points$]$ The mathematical combinations function $c(n,k)$ is usually defined in terms of factorials, as follows:

$c(n,k)=\frac{n!}{k!(n-k)!}$

The values of $c(n,k)$ can also be arranged geometrically to form a triangle in which $n$ increases as you move down the triangle and $k$ increases as you move from left to right. The resulting structure, which is called Pascal's Triangle, is arranged like this:

$c(0,0)$

$c(1,0),c(1,1)$

$c(2,0),c(2,1),c(2,2)$

$c(3,0),c(3,1),c(3,2),c(3,3)$

$c(4,0),c(4,1),c(4,2),c(4,3),c(4,4)$

Pascal's Triangle has the interesting property that every entry is the sum of the two entries above it$,$ except along the left and right edges, where the values are always $1.$ For example, $c(3,2)$ can be written as $c(2,1)+c(2,2).$ Using this fact, write a recursive implementation of the $c(n,k)$ function that uses no loops, no multiplication, and no calls to a function that computes factorial of $n.$ The function should take two input arguments n and k and return $c(n,k).$ Write a main program to display the results for $n=6$ and $k=1,2,$dots,$6.$