Let $I_{a} = [a_{1},a_{2}]$, and $I_{b} = [b_{1},b_{2}]$. With $b_{1},a_{1} \geq 0$ and $a_{2},b_{2} \leq 1$. I want to write all the cases of intervals that I can get. How many ways will I write? For example,
Case 1:$0 \leq a_{1} < a_{2} < b_{1} < b_{2} = 1$
Case 2:$0 \leq a_{1} < b_{1} < a_{2} < b_{2} = 1$
How many cases in total would you write?
I think this is a combinatorics problem. The outsides are always fixed at $1$ and $0$. Each of the inequalities can be equalities, weak, or strict (3 options for each of the 5 places). So we have the set up
$$0 < \_ < \_ < \_ < \_ < 1$$
with 15 possibilities for exchanging the inequalities to strict or weak or equality. We can fill in the blanks with the only rule being $a_1$ comes before $a_2$ and $b_1$ comes before $b_2$.
For the first blank, we have $2$ options since it must have subscript $1$.
For the second blank we have $2$ valid options since it can either be the same letter as first blank with subscript $2$ or the other letter with subscript $1$.
If the second blank has subscript $1$, then the third has $2$ options and the fourth has $1$ option. If the second blank has subscript $2$, then the third has $1$ option and the fourth has $1$ option. Either way we get that the total number of combinations remaining is $2$.
So finally we multiply to get the total number of variations on this inequality chain to be: $$ 15 \times [(2 \times 2) + 2] = 90.$$
Of those, $6$ are unique valid orderings of the letters: