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How many Topological Orderings possible?

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${\color{Green} {ABC}}{\color{Red} {DEF}}=3!3!=36$

Starting with AB or AC or BC and permuted last two elements

${\color{Green} {ABDC}}{\color{Red} {EF}} \text{and}{\color{Green} {ABDC}}{\color{Red} {FE}} =2$

${\color{Green} {ACEB}}{\color{Red} {DF}} \text{and}{\color{Green} {ACEB}}{\color{Red} {FD}} =2$

${\color{Green} {BCFA}}{\color{Red} {DE}} \text{and}{\color{Green} {BCFA}}{\color{Red} {ED}} =2$

Next 6 same as previous and just permuted first 2

${\color{Blue}{BA}\color{Green} {DC}}{\color{Red} {EF}} \text{and}{\color{Blue}{BA}\color{Green} {DC}}{\color{Red} {FE}} =2$

${\color{Blue}{CA}\color{Green} {EB}}{\color{Red} {DF}} \text{and}{\color{Blue}{CA}\color{Green} {EB}}{\color{Red} {FD}} =2$

${\color{Blue}{CB}\color{Green} {FA}}{\color{Red} {DE}} \text{and}{\color{Blue}{CB}\color{Green} {FA}}{\color{Red} {ED}} =2$

Total $48$
Answer:

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