a.) how many four-digit numbers can you form with the digits 1,2,3,4,5,6,7 if no digit is repeated? b.) how many of these four-digit numbers are odd numbers? c.) How many of them are even numbers?

### a) How many four-digit numbers can you form with the digits 1, 2, 3, 4, 5, 6, 7 if no digit is repeated?

The total number of four-digit numbers that can be formed without repetition can be calculated by considering how many choices we have for each digit. Since we are choosing from 7 distinct digits (1, 2, 3, 4, 5, 6, 7):

1. The first digit can be any of the 7 digits.
2. The second digit can be any of the remaining 6 digits.
3. The third digit can be any of the remaining 5 digits.
4. The fourth digit can be any of the remaining 4 digits.

Thus, the total number of possible four-digit numbers is:

\[
7 \times 6 \times 5 \times 4 = 840
\]

So, there are **840** different four-digit numbers that can be formed.

### b) How many of these four-digit numbers are odd numbers?

To form an odd number, the last digit must be an odd digit. The available odd digits are 1, 3, 5, and 7. Thus, there are 4 choices for the last digit.

Once the last digit (which is odd) has been chosen, we are left with 6 remaining digits for the other three places. Therefore:

1. There are 6 choices for the first digit.
2. There are 5 choices for the second digit.
3. There are 4 choices for the third digit.

Thus, the total number of odd four-digit numbers is:

\[
6 \times 5 \times 4 \times 4 = 480
\]

So, there are **480** odd four-digit numbers.

### c) How many of them are even numbers?

To form an even number, the last digit must be an even digit. The available even digits are 2, 4, and 6. Thus, there are 3 choices for the last digit.

Once the last digit (which is even) has been chosen, we are left with 6 remaining digits for the other three places. Therefore:

1. There are 6 choices for the first digit.
2. There are 5 choices for the second digit.
3. There are 4 choices for the third digit.

Thus, the total number of even four-digit numbers is:

\[
6 \times 5 \times 4 \times 3 = 360
\]

So, there are **360** even four-digit numbers.

### Summary of Answers:
a) **840** four-digit numbers can be formed.
b) **480** of these four-digit numbers are odd.
c) **360** of these four-digit numbers are even.

Would you like more details on any part of this problem?

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Here are some follow-up questions that build on this:

1. How many three-digit numbers can be formed with the digits 1, 2, 3, 4, 5, 6, 7 if no digit is repeated?
2. How many five-digit numbers can be formed with the digits 1, 2, 3, 4, 5, 6, 7 if no digit is repeated?
3. What if repetition of digits is allowed? How many four-digit numbers can be formed in this case?
4. How many four-digit numbers greater than 5000 can be formed using the digits 1, 2, 3, 4, 5, 6, 7?
5. How many odd four-digit numbers can be formed if the digit "1" must appear in every number?

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**Tip:** When solving counting problems, carefully consider restrictions on the digits, such as whether repetition is allowed and whether certain digits must occupy specific positions (e.g., odd/even requirements).

how many four-digit numbers can you form with the digits 1,2,3,4,5,6,7 if no digit is repeated? b.) how many of these four-digit numbers are odd numbers? c.) How many of them are even numbers?

Learn how to calculate the number of four-digit numbers that can be formed with the digits 1, 2, 3, 4, 5, 6, and 7 without repetition. Explore how to determine how many of these numbers are odd or even, using combinatorics and permutations. This problem is ideal for students in Grades 7-9.

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