The number of four-digit odd numbers having digits $$1, 2, 3, 4$$, each occuring exactly once, is:

A four-digit number formed from the digits $$1, 2, 3, 4$$ (each used exactly once) must satisfy two conditions:

1. All four positions (thousands, hundreds, tens, units) are filled with these digits without repetition.
2. The number is odd, so its units (last) digit must be an odd digit.

Step 1: Choose the units digit.
Among $$1, 2, 3, 4$$ the odd digits are $$1$$ and $$3$$. Hence there are $$2$$ possible choices for the units place.

Step 2: Arrange the remaining three digits.
After fixing the units digit, three distinct digits remain for the thousands, hundreds and tens places. The number of ways to arrange $$3$$ distinct objects is $$3! = 6$$.

Step 3: Total count.
Total four-digit odd numbers $$= (\text{choices for units digit}) \times (\text{arrangements of remaining digits})$$ $$= 2 \times 6 = 12$$.

Therefore, the required number of four-digit odd numbers is 12.