Determine the number of odd, nonrepeating three digits numbers that can be written using digits from the set {1, 2, 3, 4}.
I hoping this is right.
3·2·2=12
Answer 12 ways.
To determine the number of odd, nonrepeating three-digit numbers that can be written using digits from the set {1, 2, 3, 4}, we need to consider a few different factors.
First, we need to determine how many choices there are for the hundreds place digit. Since zero cannot be used as a leading digit, we cannot use the number 0 or 2 as the hundreds digit. This leaves us with two choices: 1 or 3.
Next, we need to determine how many choices there are for the tens place digit. Once we have chosen the hundreds digit, we have three digits left to choose from (since we can't repeat digits). However, we need to exclude one of the digits because it will be used as the units place digit. So, if we chose 1 as the hundreds digit, we would have three choices (2, 3, or 4) for the tens place digit. If we chose 3 as the hundreds digit, we would again have three choices (1, 2, or 4) for the tens place digit.
Finally, we need to determine how many choices there are for the units place digit. Once we have chosen the hundreds and tens place digits, only one digit is left, since we can't repeat digits.
To find the total number of odd, nonrepeating three-digit numbers, we multiply the number of choices for each place digit together.
If we chose 1 as the hundreds digit, there are 2 choices left for the tens place digit and 1 choice left for the units place digit. Therefore, there are 2 * 3 * 1 = 6 possible numbers.
If we chose 3 as the hundreds digit, there are again 2 choices left for the tens place digit and 1 choice left for the units place digit. Therefore, there are 2 * 3 * 1 = 6 possible numbers.
Adding the two cases together, we have a total of 6 + 6 = 12 odd, nonrepeating three-digit numbers that can be written using digits from the set {1, 2, 3, 4}.