Anti diagonal elements of table forming pascal triangle 
A function in $k$ and $n$ leads to the formation of this table. The elements in this table are rows of pascal triangle if we look at the anti diagonals elements of this table. They have also been colored.

According to solution $f(k,n)= C(n+k,k+1)$. 
Can anyone explain how the solution is $C(n+k,k+1)$? Here is the table.

 A: This is essentially re-indexing of the Pascal triangle: 
In order to verify that it is indeed the Pascal triangle we observe that by construction: 

  
*
  
*$f(k,n)=f(k-1,n)+f(k,n-1)$ 
  
*$f(-1, 1) = 1 = C(0,0)$
  
*$f(-1, n) = 1 = C(n-1,0)$ for all $n \in \lbrace 1,2,3 \dots \rbrace$  
  
*$f(k, 1) = 1 = C(k+1, k+1)$ for all $k \in \lbrace -1,0,1 \dots \rbrace$
  

This construction is clearly "isomorphic" to the Pascal construction with the $f(-1, n)$ entries representing the 'left leg' and the $f(k, 1)$'s representing the 'right leg' of the triangle. 
As a result we can also observe that: 


*

*Going from $f(m,n)$ to $f(m+1,n)$ is the equivalent of moving diagonally down and to the right of a Pascal triangle 

*Going from $f(k,m)$ to $f(k,m+1)$ is the equivalent of moving diagonally down and to the left of a Pascal triangle


Using any of the above observations and a simple inductive argument will give us the result. For example: 


*

*$f(0,n) = C((n-1)+1, 0+1) = C(n,1)$ (going 'down' and to the 'right')

*$f(1,n) = C(n+1, 1+1) = C(n+1,2)$

*$\vdots$

*$f(k,n) = C(n+k, k+1)$


Alternatively, 


*

*$f(k,2) = C((k+1)+1, (k+1)+0) = C(k+2,k+1)$ (going 'down' and to the 'left')

*$f(k,3) = C((k+2)+1, (k+1)+0) = C(k+3,k+1)$

*$\vdots$

*$f(k,n) = C(k+n, k+1)$


Notice how by going down and left, we are not affecting the second index of the combinations which is exactly what happens in the Pascal triangle as well. 
