erin is thinking of a 3 digit number.it uses the digits 1,7,and 4.how many numbers can you make that are even?how many can you make that are odd?explain. please answer as soon as possible!!!!!! :)

An even number ends in an even number (2, 4, 6, 8, 0)

The only three-digit numbers are 174 and 714.

An odd number ends with 1, 3, 5, 7, 9

Which numbers can you make that end in odd numbers?

I would answer it,if I knew the answer.that's why I went on to find help.I know you do t answer a question with a question.

To determine how many even and odd numbers can be made using the digits 1, 7, and 4 to form a 3-digit number:

Step 1: First, we need to determine the placement of the digits.

Since we need to form a 3-digit number, we have three places to fill: hundreds place, tens place, and ones place.

Step 2: We can use the Fundamental Counting Principle to calculate the possible combinations.

- For the hundreds place, we have 3 choices: 1, 7, or 4.
- For the tens place, we have 2 choices left, as one digit has already been used.
- For the ones place, we have 1 choice remaining.

Step 3: Now, we need to determine how many of these combinations will result in even or odd numbers.

For a number to be even, the ones place must be an even digit (0, 2, 4, 6, or 8).

- To form an even number, the ones place can only be filled with the digit 4. So, we have 1 choice.

Step 4: Calculate the number of even numbers that can be formed.

By multiplying the choices at each place together, we can calculate the total number of even numbers:

Number of even numbers = (Number of choices at hundreds place) × (Number of choices at tens place) × (Number of choices at ones place)
= 3 choices × 2 choices × 1 choice
= 6 even numbers

Therefore, there are 6 possible even numbers that can be formed using the digits 1, 7, and 4.

Step 5: Calculate the number of odd numbers that can be formed.

Since all the remaining combinations that haven't been counted as even numbers will be odd numbers:

Number of odd numbers = Total possible combinations - Number of even numbers
= 3 choices × 2 choices × 1 choice - 6 even numbers
= 6 - 6
= 0 odd numbers

Therefore, there are no possible odd numbers that can be formed using the digits 1, 7, and 4.

In summary, using the digits 1, 7, and 4, there are a total of 6 possible even numbers that can be formed, while there are no possible odd numbers.

To find out how many even and odd numbers can be made using the digits 1, 7, and 4, we need to consider the placement of the digits.

First, let's determine the number of even numbers. For a number to be even, the last digit must be even, which means it must be either 4 or 7. The first and second digits can be any of the remaining two digits.

To calculate the number of possibilities, we multiply the number of options for each digit. So for the first digit, there are three options (1, 7, or 4). For the second digit, since we have already used one digit, there are only two options left. And for the last digit, we have the choice of either 4 or 7, which gives us two options.

Therefore, the total number of even 3-digit numbers possible is 3 (options for the first digit) × 2 (options for the second digit) × 2 (options for the third digit) = 12.

Now let's determine the number of odd numbers. For a number to be odd, the last digit must be odd, which means it must be 1. The first and second digits can be any of the remaining two digits.

By following the same logic as before, we have the following: three options for the first digit, two options for the second digit, and only one option for the last digit.

Calculating these possibilities, we get 3 (options for the first digit) × 2 (options for the second digit) × 1 (option for the third digit) = 6.

Therefore, there are 12 even 3-digit numbers that can be made using the digits 1, 7, and 4, and 6 odd 3-digit numbers.

Please note that my response assumes that each digit can only be used once in forming the 3-digit number.