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P and C ... how to do?

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by praveen_gmat » Wed Aug 18, 2010 11:29 pm
A password contains at least 8 distinct digits. It takes 12 seconds to try one combination, what is the
minimum amount of time required to guarantee access to the database?
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by kmittal82 » Wed Aug 18, 2010 11:53 pm
At least 8 distinct digit means at least 8! possible passwords

12 seconds to try 1 password, hence minimum time needed = 8! x 12, whats the OA?

I think the question here is trying to trick you by using the word "combination", whereas this is really a permutation problem, since the order matters (strictly speaking all "combination locks" should be called "permutation locks" :))
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by praveen_gmat » Wed Aug 18, 2010 11:56 pm
kmittal82 wrote:At least 8 distinct digit means at least 8! possible passwords

12 seconds to try 1 password, hence minimum time needed = 8! x 12, whats the OA?

I think the question here is trying to trick you by using the word "combination", whereas this is really a permutation problem, since the order matters (strictly speaking all "combination locks" should be called "permutation locks" :))
What do you mean by OA?

The answer you have given is wrong. The answer is (10P8 + 10P9 + 10P10) × 12 seconds, though I dint understand how it is that way ..
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by kmittal82 » Thu Aug 19, 2010 12:10 am
praveen_gmat wrote:
kmittal82 wrote:At least 8 distinct digit means at least 8! possible passwords

12 seconds to try 1 password, hence minimum time needed = 8! x 12, whats the OA?

I think the question here is trying to trick you by using the word "combination", whereas this is really a permutation problem, since the order matters (strictly speaking all "combination locks" should be called "permutation locks" :))
What do you mean by OA?

The answer you have given is wrong. The answer is (10P8 + 10P9 + 10P10) × 12 seconds, though I dint understand how it is that way ..
Right, I didn't take into the account the "atleast" part of the question properly, I apologise

case (1):

exactly 8 distinct repeating digits -> #ways = 10P8

case (2)

exactly 9 distinct repeating digits -> #ways = 10P9

case (3)

exactly 10 distinct repeating digits -> #ways= 10P10

Total number of passwords to 10P8 + 10P9 + 10P10
Time taken to crack 1 password = 12 seconds

Thus, total time = (10P8 + 10P9 + 10P10)*12
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by josemaus » Thu Aug 19, 2010 9:49 am
kmittal82 wrote:
praveen_gmat wrote:
kmittal82 wrote:At least 8 distinct digit means at least 8! possible passwords

12 seconds to try 1 password, hence minimum time needed = 8! x 12, whats the OA?

I think the question here is trying to trick you by using the word "combination", whereas this is really a permutation problem, since the order matters (strictly speaking all "combination locks" should be called "permutation locks" :))
What do you mean by OA?

The answer you have given is wrong. The answer is (10P8 + 10P9 + 10P10) × 12 seconds, though I dint understand how it is that way ..
Right, I didn't take into the account the "atleast" part of the question properly, I apologise

case (1):

exactly 8 distinct repeating digits -> #ways = 10P8

case (2)

exactly 9 distinct repeating digits -> #ways = 10P9

case (3)

exactly 10 distinct repeating digits -> #ways= 10P10

Total number of passwords to 10P8 + 10P9 + 10P10
Time taken to crack 1 password = 12 seconds

Thus, total time = (10P8 + 10P9 + 10P10)*12
Are you saying that (10P* + ....) is the answer? The way I understand it, since the password is at least 8 digits long, the minimum amount of time to crack it would only need 10P8 *12
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by kmittal82 » Thu Aug 19, 2010 12:07 pm
Good point jose.

I think the "minimum" amount of time implies a guarantee that the password will be cracked.

When using the longer answer (i.e. 10P8 + ....), it guarantees that the password will be cracked, since all possible combinations will be tested.

10P8*12 is the minimum time taken only if the length of the password is 8 digits, but it won't be cracked if it contains 9 or 10 digits. Using the original answer, we cover all possible cases of the password length.
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by hero » Fri Aug 20, 2010 7:49 am
kmittal82 wrote:
praveen_gmat wrote:
kmittal82 wrote:At least 8 distinct digit means at least 8! possible passwords

12 seconds to try 1 password, hence minimum time needed = 8! x 12, whats the OA?

I think the question here is trying to trick you by using the word "combination", whereas this is really a permutation problem, since the order matters (strictly speaking all "combination locks" should be called "permutation locks" :))
What do you mean by OA?

The answer you have given is wrong. The answer is (10P8 + 10P9 + 10P10) × 12 seconds, though I dint understand how it is that way ..
Right, I didn't take into the account the "atleast" part of the question properly, I apologise

case (1):

exactly 8 distinct repeating digits -> #ways = 10P8

case (2)

exactly 9 distinct repeating digits -> #ways = 10P9

case (3)

exactly 10 distinct repeating digits -> #ways= 10P10

Total number of passwords to 10P8 + 10P9 + 10P10
Time taken to crack 1 password = 12 seconds

Thus, total time = (10P8 + 10P9 + 10P10)*12
what does the P stand for? I'm a little confused.
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by Stuart@KaplanGMAT » Fri Aug 20, 2010 10:24 am
praveen_gmat wrote:
kmittal82 wrote:At least 8 distinct digit means at least 8! possible passwords

12 seconds to try 1 password, hence minimum time needed = 8! x 12, whats the OA?

I think the question here is trying to trick you by using the word "combination", whereas this is really a permutation problem, since the order matters (strictly speaking all "combination locks" should be called "permutation locks" :))
What do you mean by OA?

The answer you have given is wrong. The answer is (10P8 + 10P9 + 10P10) × 12 seconds, though I dint understand how it is that way ..
On the actual GMAT, the answer will never say "10P8" - what's the source of this question? It's always beneficial to provide the answers choices and the source, so readers know what alternative solutions are available and whether the source is reliable.

"10P8" refers to the permutations formula:

nPk = n!/(n-k)!

in which n is the total number of objects available and k is the number of objects that you're actually arranging.

So:

for a code with 8 distinct digits, we have 10 digits available (0, 1, 2, ..., 9) and are using 8 of them;

for a code with 9 distinct digits, we have 10 digits available and are using 9 of them; and

for a code with 10 distinct digits we have 10 digits available and are using all 10 of them.

Since we want an 8 digit OR a 9 digit OR a 10 digit code, we add the individual results to get the total number of codes.

10P8 + 10P9 + 10P10

= 10!/2! + 10!/1! + 10!/0!

= 10!/2 + 10! + 10!

(0! = 1)

and, since it takes 12 seconds per try, to guarantee that we hit the code we multiply by 12 seconds:

12(10!/2 + 10! + 10!)
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