Example: How many pairs $(x,y)$ such that $x+y <= k$, where x y and k are integers and $x,y>=0, k > 0$.
My solution:
If we fix x then we can choose y in $(k - x + 1)$ ways (i.e. $y = 0,1,2,3 .... (k-x)$).
Therefore, the final answer can be calculated as follows:
$\begin{align*} &=\sum_{x = 0}^{k}\left ( k - x + 1 \right ) \\ &=\sum_{p = 1}^{k+1}\left ( p \right ) \qquad \left [ \text{ replacing k - x + 1 with p} \right ] \\ &=\frac{(k+1)(k+2)}{2} \end{align*}$
It seems that the final answer is equals to ${k+2\choose 2}\qquad$ (or choosing 2 out of k+2 distinct objects).
Question: What can be the (possible) combinatorial argument for this answer ${k+2\choose 2}$ w.r.t to the counting of $(x,y)$ pairs ?
Take $k$ items and $2$ 'dividers'. Arrange them in any order you wish. There are ${k+2}\choose{2}$ ways to do this.
The number of items (perhaps $0$) to the left of the first divider is your first addend; and the number of items between the two dividers is your second addend.
For example, with $k = 5$, you could have $|**|***$ corresponding to $0+2$. Or you could have $*|****|$ corresponding to $1+4$.