Q: How many topological orders does the following directed acyclic graph have? You may basically think of it as putting numbers on a $2\times 6$ table such that the number at the right or down is always bigger.
Of course, one may count everything manually, but I am hoping for something more elegant like using recurrence formulas or simplifying the counting process. The simplest observation one should make is that the graph has exactly one source and one sink, so at least 2 of the vertices have their position already decided in the final linear extension.
By the way, please ignore the vertices at the bottom, and labels of the vertices.
Note that as long as you know when you have an element of the top row and an element of the bottom row, the fact that the top and bottom rows are totally ordered means that that is all the information you need to completely determine the order. And at any point, you can't have used more elements from the bottom row than from the top row.
Which is to say, your problem is translatable into ordering six copies of
((representing the elements from the top row) and six copies of)(bottom row) such that the brackets are balanced. This is a problem that has been studied before, and the answer is the sixth Catalan number, $\frac17\binom{12}6=132$