Begin by constructing the number from the units digit and working up to the hundreds digit. There are two possible choices for the units digit: $3$ and $5$ (since the number must be odd). Then, after choosing that digit, there are three possible choices for the tens digit ($4$, $6$, and the remaining odd digit), since digits cannot be repeated. Finally, there are then two possible choices for the hundreds digit. Thus, there are $2 \cdot 3 \cdot 2 = \boxed{12}$ such numbers that can be formed.
How many different three-digit odd numbers can be formed using the digits 3, 4, 5 and 6 if digits cannot be repeated?
{'answer': '12', 'subject': 'Counting & Probability', 'level': 3, 'unique_id': 'test/counting_and_probability/36.json'}