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.

1st digit: 2

last digit: 3,5

So, there are 1*2*2 = 4 possibilities.

Check:
235
245
253
254

Hmmm. How did 254 make it into the list?

Ought not to be 243

To determine the number of 3-digit odd numbers less than 300 that can be formed using the integers 2, 3, 4, and 5 with no repetition of digits, we will follow these steps:

Step 1: Determine the possible choices for each digit.
In the hundreds place, we can choose any of the four integers since no repetition is allowed. Thus, we have 4 choices (2, 3, 4, 5).
In the tens place, we have again 4 choices since no repetition is allowed.
In the ones place, we can choose only an odd number because we need the final number to be odd. So, we have 2 choices (3, 5).

Step 2: Calculate the number of possibilities.
Since each digit can be chosen independently, we can multiply the number of choices in each place together to find the total number of possibilities.

Total number of possibilities = (Number of choices in the hundreds place) * (Number of choices in the tens place) * (Number of choices in the ones place)
= 4 * 4 * 2
= 32

So there are 32 possible 3-digit odd numbers less than 300 that can be formed using the integers 2, 3, 4, and 5 with no repetition of digits.