Using the integers 2,3,4,5,

How many 3-digit odd numbers less than 300 can be formed if no repetition of digits is allowed?
Please explain and show work.

Since it's an odd number LESS than 300 we know that the first number can only be 2 and therefore we have only 1 choice for the first unit. Since it's an ODD number we know that the last unit is either 3 or 5 (2 choices). No we are left with either 4 and 3 or 5 (depending on which one was chosen for the last unit), therefore we have 2 choices for the middle unit. Hence, the answer is 1x2x2 = 4 numbers.

Just for assurance, let's list out the numbers
235
245
253
243

To find the number of 3-digit odd numbers that can be formed using the integers 2, 3, 4, and 5 without repetition of digits, we need to break down the problem into steps:

Step 1: Determine the choices for the hundreds digit.
Since the numbers must be less than 300, the hundreds digit can only be either 2 or 3. So there are only two choices for the hundreds digit: 2 or 3.

Step 2: Determine the choices for the tens digit.
Since we cannot repeat any digits, we have three options remaining for the tens digit: 4, 4, and 5.

Step 3: Determine the choices for the units digit.
Since the numbers must be odd, the units digit can only be 3 or 5. So there are two choices for the units digit.

To determine the total number of possible 3-digit odd numbers, we need to multiply the number of choices at each step:

Number of choices for hundreds digit: 2
Number of choices for tens digit: 3
Number of choices for units digit: 2

Total number of possible 3-digit odd numbers = 2 * 3 * 2 = 12

Therefore, there are 12 3-digit odd numbers less than 300 that can be formed using the integers 2, 3, 4, and 5 without repetition of digits.