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Let n be 4-digit integer in which all th...

Let n be 4-digit integer in which all the digits are different. If x is the number of odd integers and y is the number of even integers, then

A

`x lt y`

B

`x gt y`

C

`x+y=4500`

D

`|x-y|=56`

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To solve the problem, we need to find the number of 4-digit integers where all digits are different, and then determine the counts of odd integers (x) and even integers (y) based on the last digit. ### Step-by-Step Solution: 1. **Understanding the Problem**: - We need to find 4-digit integers where all digits are different. - The integers can either be odd or even, depending on the last digit. 2. **Counting Odd Integers (x)**: - A 4-digit integer is odd if its last digit is one of the odd digits: 1, 3, 5, 7, or 9. - There are 5 choices for the last digit (unit place). - After choosing the last digit, we have 3 more positions (thousands, hundreds, and tens) to fill with different digits. 3. **Choosing the First Digit**: - The first digit (thousands place) cannot be 0 and cannot be the digit already used in the unit place. - Therefore, we have 8 options left for the first digit (from the digits 1-9 excluding the chosen odd digit). 4. **Choosing the Remaining Digits**: - After selecting the first digit, we have 8 digits remaining (including 0) for the hundreds place. - After selecting the hundreds place digit, we have 7 choices left for the tens place. 5. **Calculating the Total for Odd Integers**: - The total number of odd integers (x) can be calculated as: \[ x = \text{(choices for unit place)} \times \text{(choices for thousands place)} \times \text{(choices for hundreds place)} \times \text{(choices for tens place)} \] \[ x = 5 \times 8 \times 8 \times 7 = 2240 \] 6. **Counting Even Integers (y)**: - A 4-digit integer is even if its last digit is one of the even digits: 0, 2, 4, 6, or 8. - If the last digit is 0, we have 9 choices for the first digit (1-9) and then 8 for the hundreds place and 7 for the tens place: \[ \text{If last digit is 0: } 9 \times 8 \times 7 = 504 \] - If the last digit is one of the other even digits (2, 4, 6, 8), we have 4 choices for the last digit. - For the first digit, we have 8 options (1-9 excluding the chosen even digit). - For the hundreds place, we have 8 options (including 0). - For the tens place, we have 7 options. \[ \text{If last digit is 2, 4, 6, or 8: } 4 \times 8 \times 8 \times 7 = 1792 \] 7. **Calculating the Total for Even Integers**: - The total number of even integers (y) is: \[ y = 504 + 1792 = 2296 \] 8. **Comparing x and y**: - Now we compare x and y: - \( x = 2240 \) - \( y = 2296 \) - Therefore, \( x < y \). 9. **Final Results**: - We can summarize our findings: - \( x < y \) is true. - \( x + y \neq 4500 \). - \( |x - y| = 56 \) (which is not equal to 0). ### Conclusion: - The correct statements are: - \( x < y \)

To solve the problem, we need to find the number of 4-digit integers where all digits are different, and then determine the counts of odd integers (x) and even integers (y) based on the last digit. ### Step-by-Step Solution: 1. **Understanding the Problem**: - We need to find 4-digit integers where all digits are different. - The integers can either be odd or even, depending on the last digit. ...
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