Stars can fall in one of seven stellar classifications. The constellation Leo contains $9$ stars and $10$ line segments, as shown in the diagram, with stars connected by line segments having distinct stellar classifications. Let $n$ be the number of valid stellar classifications of the $9$ stars. Compute the number of positive integer divisors of $n$.
[img]https://cdn.artofproblemsolving.com/attachments/9/c/ba1cad726bc62038686a0af408e2fe60dfbde6.png[/img]
1. **Understanding the problem**: We need to find the number of valid stellar classifications for the 9 stars in the constellation Leo, where each star has a distinct classification from its connected stars. Then, we need to compute the number of positive integer divisors of this number.

2. **Analyzing the pentagon subset**: Consider the 5-star pentagon subset of the constellation, denoted as \(ABCDE\). We need to classify these stars such that no two connected stars have the same classification.

3. **Classifying \(A\) and \(B\)**: There are 7 possible classifications for \(A\) and 6 remaining classifications for \(B\), giving us \(7 \times 6 = 42\) ways to classify \(A\) and \(B\).

4. **Classifying \(D\)**:
   - **Case 1**: \(D\) has the same classification as \(A\) or \(B\). There are 2 choices for \(D\) (same as \(A\) or \(B\)), 6 choices for \(C\) (different from \(A\) and \(B\)), and 5 choices for \(E\) (different from \(A\), \(B\), and \(C\)). This gives \(2 \times 6 \times 5 = 60\) ways.
   - **Case 2**: \(D\) has a different classification from \(A\) and \(B\). There are 5 choices for \(D\), 5 choices for \(C\) (different from \(A\), \(B\), and \(D\)), and 5 choices for \(E\) (different from \(A\), \(B\), \(C\), and \(D\)). This gives \(5 \times 5 \times 5 = 125\) ways.

5. **Total classifications for the pentagon**: Summing the two cases, we get \(60 + 125 = 185\) ways to classify the pentagon given the classifications of \(A\) and \(B\). Therefore, the total number of valid classifications for the pentagon is \(42 \times 185 = 7770\).

6. **Classifying the remaining 4 stars**: Each of the remaining 4 stars can be classified in 6 ways (since they are connected to at most 3 other stars, and there are 7 classifications in total). Thus, the total number of valid classifications for the entire constellation is:
   \[
   7770 \times 6^4 = 7770 \times 1296 = 10077120
   \]

7. **Finding the number of positive integer divisors**: To find the number of positive integer divisors of \(10077120\), we need its prime factorization:
   \[
   10077120 = 2^7 \times 3^4 \times 5^1 \times 7^1
   \]
   The number of positive integer divisors is given by multiplying the incremented exponents:
   \[
   (7+1)(4+1)(1+1)(1+1) = 8 \times 5 \times 2 \times 2 = 160
   \]

The final answer is \(\boxed{160}\)