The problem defines a 'frabjous' number as a 3-digit number in which all digits are odd, and no two adjacent digits are the same. The possible odd digits are 1, 3, 5, 7, and 9. We will calculate the total number of such 3-digit frabjous numbers by considering the following conditions:
- For the first digit: Since the number is a 3-digit number, the first digit can be any of the 5 odd digits: \{1, 3, 5, 7, 9\}. So, there are 5 choices for the first digit.
- For the second digit: The second digit must also be odd, but it cannot be the same as the first digit. Hence, there are only 4 choices for the second digit because one odd digit is already taken by the first digit.
- For the third digit: The third digit also needs to be odd, and it cannot be the same as the second digit. So, there are again 4 choices for the third digit.
Thus, the total number of frabjous numbers is the product of the number of choices for each digit:
\[
5 \times 4 \times 4 = 80
\]
Therefore, the total number of frabjous numbers is \( 80 \).