The total count of curricula: \({8 \choose 4}\cdot{10 \choose 3}\)

To handle the prerequisites we need to consider the problem case by case. It needs some care to organize the cases. Let’s agree on a scheme where a 3-tuple of 0’s and 1’s represents the absence and presence of low-level courses from 1 to 3 in the proposed curriculum. There are 8 logically possible cases, but not all of them are legitimate. Because, if \(L_1\) is not in, then \(L_2\) and \(L_3\) must be, otherwise there will be no higher level courses. This leaves us 5 cases to consider.

Case Count
\((1,1,1)\) \({5 \choose 1}\cdot {10 \choose 3}\)
\((1,1,0)\) \({5 \choose 2}\cdot {5 \choose 3}\)
\((1,0,1)\) \({5 \choose 2}\cdot {5 \choose 3}\)
\((1,0,0)\) \({5 \choose 3}\cdot {5 \choose 3}\)
\((0,1,1)\) \({5 \choose 2}\cdot {5 \choose 3}\)

The answer is the sum of the counts.