Math Problem Statement

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?

Solution

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:

7×6×5×4=8407 \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:

6×5×4×4=4806 \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:

6×5×4×3=3606 \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?


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?

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).

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Math Problem Analysis

Mathematical Concepts

Combinatorics
Permutations
Odd and Even Numbers

Formulas

Total permutations for four-digit numbers: 7 × 6 × 5 × 4
Permutations for odd numbers: 6 × 5 × 4 × 4
Permutations for even numbers: 6 × 5 × 4 × 3

Theorems

Basic Principle of Counting
Permutation without Repetition

Suitable Grade Level

Grades 7-9