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Ok in a normal set of N numbers, I might take the length of the number set, like 4 numbers long, raise that to the power of the base numbering scheme 0-9 so 10^4th=10000 and subtract 1. I get that.

Suppose I cannot reuse any number in the set. 1-9 I assume my base is 8. But i cant do 111111111... 112345678 or anything like that.

what method would I use to figure out the maximum number of permutations between 123456789 and 987654321?

oh and all the permutations must be 9 digits.

I have a guess, that it might be (8^9)/8 but I'm not sure that makes sense.

kimchi lover
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j0h
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  • the number of permutations in $n!$ simply by induction fix one digit $n$ choices, and permute all the other $(n-1)!$. – zwim Sep 15 '22 at 19:23

1 Answers1

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Suppose I cannot reuse any number in the set. 1-9 I assume my base is 8. But i cant do 111111111... 112345678 or anything like that.

I believe the base would be 9. (Count the number of integers between 1-9)

what method would I use to figure out the maximum number of permutations between 123456789 and 987654321?

I believe we could rephrase this question to "Total number of 9 digit integers with no repeated digits" That is because we are finding all the 9 digit numbers between the lowest 9 digit number (123456789) and the greatest 9 digit number (987654321). The answer to this would be P(9, 9) (which is 9!). We are permuting 9 distinct items (unique digits) into 9 spots.

A more intuitive approach might be to draw out the spots and fill them in with the number of choices we have. Imagine we have 9 empty spots and we fill each with a digit:

How many options do we have for the first spot? Well, we have 9 digits to pick from so 9.

9 _ _ _ _ _ _ _ _

How about for the second spot? We already chose a digit to put in the first spot, and because we can't repeat we have just 8 options left.

9*8 _ _ _ _ _ _ _ following this logic until all spots are filled:

$9\times8\times7\times6\times5\times4\times3\times2\times1= 9! = 362880$

Hope this helped.

shrizzy
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