How many two digit numbers have exactly $4$ positive factors? $($Here $1$ and  the number $n$ are also considered as factors of $n. )$
To determine how many two-digit numbers have exactly 4 positive factors, we need to understand the structure of numbers with exactly 4 factors. 

A number \( n \) has exactly 4 positive factors if and only if it can be expressed in one of the following forms:
1. \( n = p^3 \), where \( p \) is a prime number.
2. \( n = p \cdot q \), where \( p \) and \( q \) are distinct prime numbers.

Let's analyze each case separately:

1. **Case 1: \( n = p^3 \)**
   - We need to find prime numbers \( p \) such that \( p^3 \) is a two-digit number.
   - The prime numbers \( p \) that satisfy \( 10 \leq p^3 < 100 \) are:
     - \( 2^3 = 8 \) (not a two-digit number)
     - \( 3^3 = 27 \)
     - \( 5^3 = 125 \) (not a two-digit number)
     - \( 7^3 = 343 \) (not a two-digit number)
   - Therefore, the only valid \( p \) is \( 3 \), giving us one number: \( 27 \).

2. **Case 2: \( n = p \cdot q \)**
   - We need to find pairs of distinct prime numbers \( p \) and \( q \) such that \( 10 \leq p \cdot q < 100 \).
   - Let's list the prime numbers less than 50 (since \( p \cdot q \) must be less than 100 and both \( p \) and \( q \) must be distinct primes):
     - \( 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47 \)
   - We will now find all pairs \( (p, q) \) such that \( 10 \leq p \cdot q < 100 \):
     - For \( p = 2 \): \( 2 \cdot 3 = 6 \) (not valid), \( 2 \cdot 5 = 10 \), \( 2 \cdot 7 = 14 \), \( 2 \cdot 11 = 22 \), \( 2 \cdot 13 = 26 \), \( 2 \cdot 17 = 34 \), \( 2 \cdot 19 = 38 \), \( 2 \cdot 23 = 46 \), \( 2 \cdot 29 = 58 \), \( 2 \cdot 31 = 62 \), \( 2 \cdot 37 = 74 \), \( 2 \cdot 41 = 82 \), \( 2 \cdot 43 = 86 \), \( 2 \cdot 47 = 94 \) (14 valid pairs)
     - For \( p = 3 \): \( 3 \cdot 5 = 15 \), \( 3 \cdot 7 = 21 \), \( 3 \cdot 11 = 33 \), \( 3 \cdot 13 = 39 \), \( 3 \cdot 17 = 51 \), \( 3 \cdot 19 = 57 \), \( 3 \cdot 23 = 69 \), \( 3 \cdot 29 = 87 \), \( 3 \cdot 31 = 93 \) (9 valid pairs)
     - For \( p = 5 \): \( 5 \cdot 7 = 35 \), \( 5 \cdot 11 = 55 \), \( 5 \cdot 13 = 65 \), \( 5 \cdot 17 = 85 \), \( 5 \cdot 19 = 95 \) (5 valid pairs)
     - For \( p = 7 \): \( 7 \cdot 11 = 77 \), \( 7 \cdot 13 = 91 \) (2 valid pairs)
     - For \( p = 11 \) and higher primes, \( p \cdot q \) will exceed 100 for any \( q \geq p \).

Summing up the valid pairs:
\[ 14 + 9 + 5 + 2 = 30 \]

Combining both cases, we have:
\[ 1 \text{ (from } p^3 \text{)} + 30 \text{ (from } p \cdot q \text{)} = 31 \]

The final answer is \(\boxed{31}\)