How many 3-digit odd numbers can be formed using the digits 1, 2, 3, 4, 5, 6 if repetition of digits is not allowed?

For an odd number, the units digit must be one of 1, 3, or 5, so there are 3 choices. After choosing that, the hundreds place can be filled by any of the remaining 5 digits, and the tens place by any of the remaining 4 digits. Hence the count is $3\times 5\times 4=60$.