MHB Exploring the Possibilities: Solving an 8-Digit Even Number Permutation Question

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To determine how many eight-digit even numbers can be formed using the digits 7, 5, 4, 5, 7, 5, 0, and 7, the last digit must be 0 to ensure the number is even. The first digit can only be 7, 5, or 4, as starting with 0 would not yield an eight-digit number. The discussion emphasizes breaking the problem down step by step, considering the placement of each digit. Participants suggest analyzing combinations for the first digit and then calculating permutations for the remaining digits. A structured approach is essential for arriving at the correct total of valid permutations.
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It's a permutation question, so I don't know where else to post this.

How many eight digit even numbers are possible with the digits 7 5 4 5 7 5 0 7?

Please explain step by step.
 
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Raerin said:
It's a permutation question, so I don't know where else to post this.

How many eight digit even numbers are possible with the digits 7 5 4 5 7 5 0 7?

Please explain step by step.

Welcome to MHB Raerin! :)

Can you give an indication what you have tried or considered?
That helps us to understand how we can best help you.
Can you for instance give a couple of examples of numbers that satisfy the criteria?
 
Just to get you started, with questions asking 'how many ways?' it's often a help to take one digit at a time (first, second, third...)

For the first digit, would you agree it can only be a 7, 5 or 4? If it were a zero, this would not be an eight digit number. So, there are 3 combinations of the first digit.

Can you see now how to work toward a solution?
 

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