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This seems to be a pretty tough question. Seven digit numbers range from 1,000,000 to 9,999,999. There are 9,000,000 of them. The number of these divisible by 125 will be 9,000,000/125 which is 72,000.

If no number contains a zero, and all the digits are different, then we know two things:

- Since multiples of 125 will end in 5 or 0, our numbers will all end in 5.

- The other 6 digits must be chosen the eight digits 1,2,3,4,6,7,8,9.

How many ways can you choose 6 things out of 8? This is 8C6. The number of arrangements of these 6 things would be 8P6. How are you doing so far - if you're with me then resubmit your question starting with "Please continue to find the number ..."

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Q: Find the number of seven-digit numbers such that every number is divisible by 125 no number contains a zero and no digit appears more that once in the number?
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