Here’s this week’s practice question (regular multiple choice, just one answer), we’ll be posting the answer tomorrow. Good luck!

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?

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There are totally 10 digits that can be chosen without repetition from 0-9. So for a 9 digit password we can choose 9 digits from 10 choices so
10P9
and for a 10 digit password we can choose 10 digits from a list of 10 digits so
10P10
This implies 10P9 + 10P10 = 10! + 10!(since 10P9 and 10P10 are both equal to 10!)
hence we get 2*10!

We have total 10 numbers (0,1,2,….9)
For nine digit password
Here first digit can be filled by 10C1 ways. After filling that we have (10-1) =9 numbers.
So, Second digit can be filled by 9C1 ways. Similarly,
Third digit can be filled by 8C1 ways and so on.

“At least 9 digits long” is the key phrase. This means the password can be 9 digits or 10 digits long (since we can have max ten digits 0-9). If 9 digits long and no repetitions , 10*9*8*7*6*5*4*3*2=10! passwords possible. If 10 digits long and no repetitions , 10*9*8*7*6*5*4*3*2*1=10! passwords possible. So, in total 2*10! passwords are possible.

A. nine digit passwords+ten digit passwords
choose 9 digits one way 10C9. out of those 9!. ||ly 10C10 out of those 10!

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can the zero can be placed inthe first place, as zero at the first place does not plays the value

Yes, you can have ‘0’ in the first place, because we are not dealing with a number but with a passcode. Hope that makes sense :).

The answer would be

B. 2*10!

i.e.,10!+10!

There are totally 10 digits that can be chosen without repetition from 0-9. So for a 9 digit password we can choose 9 digits from 10 choices so

10P9

and for a 10 digit password we can choose 10 digits from a list of 10 digits so

10P10

This implies 10P9 + 10P10 = 10! + 10!(since 10P9 and 10P10 are both equal to 10!)

hence we get 2*10!

We have total 10 numbers (0,1,2,….9)

For nine digit password

Here first digit can be filled by 10C1 ways. After filling that we have (10-1) =9 numbers.

So, Second digit can be filled by 9C1 ways. Similarly,

Third digit can be filled by 8C1 ways and so on.

(10C1) *(9C1) *(8C1) *(7C1) *(6C1) *(5C1) *(4C1) *(3C1) *(2C1)

=10*9*8*7*6*5*4*3*2

=10!

For ten digit password, on the same way

(10C1) *(9C1) *(8C1) *(7C1) *(6C1) *(5C1) *(4C1) *(3C1) *(2C1) *(1C1)

=10*9*8*7*6*5*4*3*2*1

=10!

So, Total combination =10!+10!=2*10!

Answer is B.

“At least 9 digits long” is the key phrase. This means the password can be 9 digits or 10 digits long (since we can have max ten digits 0-9). If 9 digits long and no repetitions , 10*9*8*7*6*5*4*3*2=10! passwords possible. If 10 digits long and no repetitions , 10*9*8*7*6*5*4*3*2*1=10! passwords possible. So, in total 2*10! passwords are possible.

A. nine digit passwords+ten digit passwords

choose 9 digits one way 10C9. out of those 9!. ||ly 10C10 out of those 10!