The number of positive integers less than 50, having exactly two distinct factors other than 1 and itself, is

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To find the positive integers less than 50 with exactly two distinct factors other than 1 and itself:

1. **Case I: Perfect cubes below 50**  
   Perfect cubes are numbers of the form $$ p^3 $$, where $$ p $$ is a prime number.  
   Below 50, the perfect cubes are:  
   $$ 2^3 = 8 $$ and $$ 3^3 = 27 $$.  
   Thus, there are **2 numbers** in this case.

2. **Case II: Product of two distinct primes**  
   For the product of two primes to be less than 50, both primes must be below 25.  
   The primes less than 25 are:  
   $$ 2, 3, 5, 7, 11, 13, 17, 19, 23 $$.

- **Sub-case A: Both primes are less than $$ \sqrt{50} $$**  
     Primes less than $$ \sqrt{50} $$ ($$ \approx 7.07 $$) are $$ 2, 3, 5, 7 $$.  
     Any product of two numbers among these yields a product less than 50.  
     The number of pairs is given by:  
     $$ ^4C_2 = \frac{4!}{2!(4-2)!} = 6 $$.

- **Sub-case B: Both primes are greater than $$ \sqrt{50} $$**  
     Primes greater than $$ \sqrt{50} $$ are $$ 11, 13, 17, 19, 23 $$.  
     Any product of two numbers among these yields a product greater than or equal to 50.  
     Hence, there are **0 pairs** in this sub-case.

- **Sub-case C: One prime is less than $$ \sqrt{50} $$, and one is greater**  
     Between the two lists ($$ \sqrt{50}: $$), we calculate pairs:  
     - $$  $$ and $$  $$ can pair with $$  $$ and $$  $$.  
     - Other primes ($$ ,, $$) can only pair with $$  $$.

Total pairs = **7 pairs**.

3. **Total Count:**  
   Adding all cases together:  
   $$ 2 + 6 + 0 + 7 = 15. $$

Thus, there are **15 positive integers** less than 50 with exactly two distinct factors other than 1 and itself.