What are the total number of ways in which $i$, $j$ can be chosen subject to constrain $1\leq i \leq j \leq n$? What are the total number of ways in which $i$,$j$ can be chosen subject to constrain $1\leq i \leq j \leq n$ ? All are integers. My progress is: I believe that out of the $n$ entries, there are $n \choose 2$ ways to choose $i,j$. But, the given answer is ${n \choose 2} + n$. Some explanation would be helpful.

We want to find the number of ways we can choose integers $i, j$ such that $1 \leq i \leq j \leq n$. There are two possibilities:

1. $i < j$: The number of such selections is the number of two element subsets $\\{1, 2, 3, \ldots, n\\}$ since the smaller number we select must be $i$ and the larger one must be $j$. The number of such subsets is $$\binom{n}{2}$$

2. $i = j$: The number of such selections is the number of ways we can select one element from the set $\\{1, 2, 3, \ldots, n\\}$ since the selected number must equal both $i$ and $j$. The number of ways we can do this is $n$.




Since these cases are mutually exclusive and exhaustive, the number of ways we can choose integers $i, j$ such that $1 \leq i \leq j \leq n$ is $$\binom{n}{2} + n$$