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GRE Permutations: Practice Question of the Week #37

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?

1. 9! + 10!
2. 2 x 10!
3. 9! x 10!
4. 19!
5. 20!

6 Responses to GRE Permutations: Practice Question of the Week #37

1. ashok July 16, 2012 at 4:52 am #

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

• Chris July 18, 2012 at 4:25 pm #

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 :).

2. madhan_2005 February 7, 2012 at 4:58 am #

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!

3. Rubaet February 6, 2012 at 10:07 pm #

We have total 10 numbers (0,1,2,….9)
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!

4. Swagoto February 6, 2012 at 6:39 pm #

“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.

5. m February 6, 2012 at 5:46 pm #