Hey guys! Today, let's dive into a fun math problem that involves counting. We're going to figure out how many 4-digit numbers we can create where no digit is repeated, and the number is even. This is a classic problem that combines the principles of permutations and the divisibility rule for even numbers. So, buckle up and let's get started!
Understanding the Basics
Before we jump into the solution, let's make sure we understand the key concepts involved. First, we need to know what a 4-digit number is. A 4-digit number is any number from 1000 to 9999. Next, we need to understand what it means for a number to be even. An even number is any integer that is divisible by 2. In simpler terms, it's a number that ends in 0, 2, 4, 6, or 8. Lastly, the phrase "no digit repeats" means that each digit in the number must be unique. For example, 1234 is a valid number, but 1223 is not because the digit 2 is repeated.
Breaking Down the Problem
To solve this problem, we'll use a step-by-step approach. We'll consider each digit place (thousands, hundreds, tens, and ones) and determine the number of choices we have for each place, keeping in mind the constraints of the problem. The constraints are:
- The number must be a 4-digit number.
- No digit can be repeated.
- The number must be even.
The Importance of Being Even
The even number constraint is crucial because it limits the choices for the last digit (the ones place). This is where we'll start our calculation. The ones place can only be filled by an even digit (0, 2, 4, 6, or 8). The number of choices for the ones place will affect the choices for the other places.
Step-by-Step Solution
Let's break down the problem step by step to make it easier to understand. We'll start by considering the possible digits for each place value, keeping in mind the rules we discussed earlier.
1. The Ones Place
As we mentioned, the ones place is the most restricted because the number has to be even. The possible digits for the ones place are 0, 2, 4, 6, and 8. However, we need to consider two cases:
- Case 1: The ones place is 0.
- Case 2: The ones place is not 0 (i.e., 2, 4, 6, or 8).
We'll handle these cases separately because they have different implications for the other digits.
Case 1: Ones Place is 0
If the ones place is 0, we have only 1 choice for this position. This simplifies the problem because 0 doesn't affect the choices for the thousands place (which cannot be 0 in a 4-digit number).
- Ones place: 1 choice (0)
Now, let's move to the other digits.
- Thousands place: Since we've used 0, we have 9 remaining digits (1 through 9) to choose from. So, there are 9 choices for the thousands place.
- Hundreds place: We've used two digits (one for the ones place and one for the thousands place), so we have 8 remaining digits to choose from. Therefore, there are 8 choices for the hundreds place.
- Tens place: We've now used three digits, leaving us with 7 remaining digits. So, there are 7 choices for the tens place.
To find the total number of even numbers in this case, we multiply the number of choices for each place:
1 (ones) * 9 (thousands) * 8 (hundreds) * 7 (tens) = 504
So, there are 504 even numbers that end in 0.
Case 2: Ones Place is Not 0
Now, let's consider the case where the ones place is not 0. This means the ones place can be 2, 4, 6, or 8. So, we have 4 choices for the ones place.
- Ones place: 4 choices (2, 4, 6, or 8)
This case is a bit trickier because the thousands place cannot be 0, and we've already used one non-zero digit for the ones place.
- Thousands place: Since we can't use 0 and we've used one digit for the ones place, we have 8 choices left (1 through 9, excluding the digit used in the ones place). So, there are 8 choices for the thousands place.
- Hundreds place: We've used two digits (one for the ones place and one for the thousands place), so we have 8 remaining digits to choose from (including 0). Therefore, there are 8 choices for the hundreds place.
- Tens place: We've now used three digits, leaving us with 7 remaining digits. So, there are 7 choices for the tens place.
To find the total number of even numbers in this case, we multiply the number of choices for each place:
4 (ones) * 8 (thousands) * 8 (hundreds) * 7 (tens) = 1792
So, there are 1792 even numbers that do not end in 0.
Combining the Cases
To find the total number of 4-digit even numbers with no repeating digits, we need to add the results from both cases:
Total = Case 1 + Case 2
Total = 504 + 1792 = 2296
Therefore, there are 2296 such numbers.
Final Answer
The final answer is 2296, which corresponds to option (a).
Why This Method Works
This step-by-step method works because it systematically considers each constraint. By starting with the most restrictive condition (the ones place being even), we reduce the complexity of the problem. Breaking the problem into cases further simplifies the calculation by handling the special condition of 0 separately.
Key Takeaways
- Break down complex problems: Divide the problem into smaller, manageable steps.
- Identify constraints: Understand the limitations and conditions of the problem.
- Consider cases: If there are special conditions, handle them separately.
- Multiply choices: Use the counting principle to multiply the number of choices for each step.
Common Mistakes to Avoid
- Forgetting the zero: The digit 0 can be tricky because it cannot be the first digit of a 4-digit number. Remember to account for this when calculating the choices for each place.
- Not considering cases: Failing to separate the cases where the ones place is 0 and not 0 can lead to an incorrect answer.
- Double-counting: Make sure you're not counting the same number multiple times. This is why it's crucial to ensure no digits are repeated.
Practice Problems
To solidify your understanding, try solving similar problems. Here are a few examples:
- How many 3-digit numbers are there where no digit repeats and the number is odd?
- How many 5-digit numbers can be formed using the digits 1 to 9 if no digit is repeated?
- How many 4-digit numbers are there where the number is divisible by 5 and no digit repeats?
Conclusion
Calculating the number of 4-digit even numbers with no repeating digits might seem challenging at first, but by breaking it down into smaller steps and considering the constraints, it becomes quite manageable. Remember, the key is to understand the problem, identify the constraints, and systematically count the possibilities. Keep practicing, and you'll become a pro at these types of problems in no time!
I hope this explanation was helpful, guys. If you have any questions or want to explore more math problems, feel free to ask. Happy calculating!
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