## § Intuition for why choosing closed-closed intervals of `[1..n]`

is $(n+1)C2$

- $nC2$ counts all intervals $\{ [i, j]: i > j \}$.
- To count intervals $[i, i]$, there are $n$ of them, so it's $nC2 + n$ which is $n(n-1)/2 + n$, which is $n(n+1)/2$ or $(n+1)C2$.
- Combinatorially, add a "special point
`*`

" to `[1..n]`

. If we pick a pair `(i, *)`

from the $(n+1)C2$, take this to mean that we are picking the interval `[i, i]`

.