To solve this, we need to determine how many 8-digit odd phone numbers exist such that all their digits are in ascending order and without repetition, with none of them starting with zero.

Steps:

1. Understand the structure of the phone number:

   • The number is odd and consists of 8 digits.
   • None of the digits can start with zero.
   • The digits are arranged in ascending order without repetition.
   • The last digit must be odd because the number is odd.

2. Digits available:

   • The digits we can use are 1 through 9 (since zero is not allowed at the start, and repetition is not allowed).
   • Among these, the odd digits are 1, 3, 5, 7, and 9.

3. The last digit must be odd:

   • Since the number must end in an odd digit, the last digit has to be chosen from {1, 3, 5, 7, 9}.

4. Choose the remaining 7 digits:

   • Once the last (odd) digit is chosen, the remaining 7 digits must be selected from the digits smaller than that last digit.
   • For example, if the last digit is 9, we need to pick 7 digits from {1, 2, 3, 4, 5, 6, 7, 8}.
   • If the last digit is 7, we can only pick from {1, 2, 3, 4, 5, 6}.

5. Summing up possibilities:

   • For each possible odd last digit, count how many ways we can select 7 digits from the available smaller digits:
     • If the last digit is 9, we choose 7 digits from {1, 2, 3, 4, 5, 6, 7, 8}. This is $\binom{8}{7} = 8$ ways.
     • If the last digit is 7, we choose 7 digits from {1, 2, 3, 4, 5, 6}. This is $\binom{6}{7} = 1$ way.
     • If the last digit is 5, we choose 7 digits from {1, 2, 3, 4}. This is ( \bin bus