Note that we are told that no digit may be repeated.

There are total of 10 digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9.

As password must be at least 9 digits long then it can consist of either 9 or all 10 digits.

If password consists of 9 digits then # of passwords possible is \(C^9_{10}*9!=10*9!=10!\): \(C^9_{10}\) - # of ways to choose 9 digits which will be used in password and \(9!\) arranging these digits (or in another way \(P^9_{10}=10!\) - choosing 9 digits out of 10 when the order matters);

If password consists of all 10 digits then # of passwords possible is simply \(10!\);

So total # of passwords possible is \(10!+10!=2*10!\).

Answer: B.
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why 9! is multiplied with 10 ?

thanks
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It's good to realize the permutation formula: n! / (n-k)!

We then got:

10! / (10-9)! = 10!/1! = 10!
and
10! / (10-10)! = 10!/0! = 10! = 10!/1! = 10! (

The answer is than: 2 * 10!
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The question must mention that the password can start with a '0'. Bunuel, if my reasoning is correct, would you please edit the question?
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9 digit passwords = \(10_C_9\) * 9! = 10!
10 digit passwords = \(10_C_{10}\) * 10!= 10!
sum = 2*10!
Answer B

The number of ways to create a 10-digit password from 10 digits is 10P10 = 10!.

Thus, the number of ways to create the codes is 10! + 10! = 2(10!).

Answer: B
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