Discussion in 'Algebra' started by nycmathguy, May 19, 2022.

1. ### nycmathguy

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A popular website requires users to create a password consisting of digits only. If no digit may be repeated and each password must be at least 9 digits long, how many passwords are possible?
A. 9! + 10! B. 2 × 10! C. 9! × 10! D. 19! E. 20!

Here is the explanation I found online. For this thread, I want you to explain the explanation, if you will, that I found online.

EXPLANATION

If we choose the 10 different digits then they can be arranged (permutations) in 10! ways. But the question asks at least 9 digits. So we have the possibility of choosing only 9 digits for the password ( but digit shouldn't repeat), so we can have a total of 10 different combinations and each combination can be arranged in 9! ways. Therefore 10 x 9! + 10! = 10! + 10! = 2 x 10!

nycmathguy, May 19, 2022

2. ### MathLover1

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correct

MathLover1, May 19, 2022
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3. ### nycmathguy

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This is not my explanation. I asked for you to explain the explanation I found online. The explanation itself makes zero sense to me as written. You say?

nycmathguy, May 19, 2022
4. ### MathLover1

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note: each password must be at least 9 digits long -> since there are 10 digits in all, means password could be 9 digits long or 10 digits long

Since there are 10 digits (0, 1, 2, 3, 4, 5, 6, 7, 8, 9) and no digit may be repeated then the maximum length of the password is 10.
That is, with 11 digits or more you would repeat at least one digit.

So, we have to compute the number of passwords with 9 digits and the number of passwords with 10 digits, and sum those numbers.

10 digits:
This is very easy, it’s just 10! That is, permutations of the 10 digits.

9 digits:
Permutations of 10 digits taken 9 at a time.
Perm(10, 9) = 10! / 1! = 10!

Total: 2*10! = 7257600

MathLover1, May 19, 2022
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