I see. This "at least" changes everything, it means that password can have 8, 9 or 10 digits (more than 10 is not possible as per stem digits must be distinct).

Total # of passwords possible for 8 digits is \(P^8_{10}=\frac{10!}{2}\);
Total # of passwords possible for 9 digits is \(P^9_{10}=10!\);
Total # of passwords possible for 10 digits is \(P^{10}_{10}=10!\).

Time needed to guarantee access to database is \((\frac{10!}{2}+10!+10!)*\frac{1}{5}=\frac{10!}{2}\) minutes.

Answer: D.
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Hi, and welcome to the Gmat Club. Below is the solution for your problem.

Total # of passwords possible is \(P^8_{10}=10*9*8*7*6*5*4*3\) (picking 8 different digits out of ten when order of the digits matters, or another approach 10 choices for the 1st digit, 9 choices for the second and so on), 1/5 minute (12 seconds) for one combination, means that time needed to guarantee access to database is \(\frac{P^8_{10}}{5}\) minute.

Hope it's clear.
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Thank you - But the answer choices are as follows -

a) 8!/5 b) 8!/2 c) 8! d) 10!/2 and e) 5/2.10!

Answer to the question you've posted is \(\frac{P^8_{10}}{5}\), which can also be written as \(\frac{10!}{10}=9!\). But still no match with any of the answer choices. Pleas check the question (maybe some info from the stem is missing) or site the source.
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Source is Veritas material

I did check the question and following info is missing - If the password is known to consist of at least 8 digits - "at least" is missing

Bunuel,
I used combination for this problem since the qs uses the word comb. I thought when the qs doesn't indicate anything about the order then we have to assume that it doesn't matter. Please correct me.

Most combinatorics questions can be solved with different approaches. Did you get correct answer with your approach? If yes, then it doesn't matter which formula you used.

Next, it's a password, of course order matters: 12345678 is different from 87654321. \(P^8_{10}\) directly gives the total # of 8 digit passwords (when digits are distinct), because it counts every case of 8 digits possible from 10 (eg the case of "12345678") as many times as the digits can be arranged there (87654321, 56781234, ...). Different way to get the same answer: \(C^8_{10}\) # of ways to choose 8 digits out of 10 multiplied by 8! to get different arrangements of these digits, so \(C^8_{10}*8!\), which is the same as \(P^8_{10}\) (\(C^8_{10}*8!=P^8_{10}\)).

Combinatorics questions are often confusing for many and there are numerous resources with really strange (not to say more) advices, like "if there is some specific word in stem (eg "combination") then use C" or "if there is a direct indication that order matters then use P" and so on.

I think one should understand the formula and not just memorize it. When solving the question we should first understand what it is about and only after that choose the proper tool (formula) and not first choose formula (based on one word) and then trying to squeeze numbers in it.

Hope it helps.
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Could you please correct your first port ? As you said, the stem is missing the "at least" and you forgot to put the answer choices.

Thanks.

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?

The questions seems awkward...
Anyways, I will consider that password consists of 8 distinct digits (and of 8 numbers only).
So there will be 8! different passwords and hence the minimum time to access the database is 8! * 12.

Key word here is AT LEAST 8 digits,

case (1): When we have exactly 8 digits in the password

Then the number of passwords = \(10_P_8\)

case (2) : When we have 9 digits in the password

Then the number of passwords = \(10_P_9\)

case (3) : When we have all the digits i.e 10 digits in the password

Then the number of passwords = \(10_P_{10}\)

Total number of passwords to\(10_P_8 + 10_P_9 + 10_P_{10}\)

Time taken to crack 1 password \(= 12 seconds\)

Thus, total time \(= (10_P_8 + 10_P_9 + 10_P_{10})*12\)

Hi bunuel,


In this question we consider 10 digits 0-9

but for a 8 digit number, the first digit cannot be 0 so there would be only 9 ways to select the first digit.

Have we missed that here??
Please explain...

It is a password which consists of at least 8 digits, not an eight-digit number, so it can also start with a 0.
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I think the question is flawed.
Let me chalk out the conditions according to the question's language.

Condition 1 - Password contains at least 8 digits
Condition 2 - All the digits of the password are different.

For both the conditions to be satisfied, there can be three cases.

Case 1 - 8 distinct digits. Bunuel has solved this part.

Case 2 - 9 distinct digits. This will be given by

\(10*9*8*.....*2\) = \(10!\)

To try each permutation 1/5 of a minute is needed.

Case 3 - 10 distinct digits. Same as Case 2.

So the worst case scenario for our "testing machine" will be

8 digits testing - No joy!
9 digits testing - No joy!
10 digits testing - exactly the last one tested turns out to be the match.

Bunuel Please see.

I may be overthinking too much about this!

But as we need to solve a certain password should not we take the no. of highest possibility that is 10! ??!!!! Plz explain

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Plz give kudos, buddies!!!!

Total distint digits in one password: Minimum 8 & Maximum 10.

For 8 digits password, no. of distinct combinations possible= 8!*10C8= 10!/2!= 10!/2

For 9 digits password, no. of distinct combinations possible= 9!*10C9= 10!/1!= 10!

For 10 digits password, no. of distinct combinations possible= 10!*10C10= 10!/0!= 10!

Total possible passwords=
10!*0.5 + 10! + 10!= 10!*2.5

Time required to try one password= 12sec= 1/5 min.

Therefore, Time required to try 10!*2.5 password= 10!*2.5/5 min= 10!/2 mins.

Hence, Ans D

OA:D
There are \(10\) possibilities for \(1\) st digit i.e \(1\) st digit can be \(0,1,2,3,4,5,6,7,8,9\).

After the \(1\) st digit, the number of possibilities keeps decreasing by \(1\) as digits cannot be repeated.

Number of possibilities when there are \(8\) digits in the password: \(10*9*8*7*6*5*4*3=\frac{10!}{2}\)

Number of possibilities when there are \(9\) digits in the password : \(10*9*8*7*6*5*4*3*2 =\frac{10!}{1}\)

Number of possibilities when there are \(10\) digits in the password : \(10*9*8*7*6*5*4*3*2*1 = 10!\)

Total Number of possibilities : \(\frac{10!}{2} +10!+10!=\frac{5*10!}{2}\)

Time taken for trying \(1\) password: \(12\) seconds \(= \frac{12}{60}\quad minutes=\frac{1}{5}\quad minutes\)

Time taken for trying \(\frac{5*10!}{2}\) passwords \(= \frac{1}{5}*\frac{5*10!}{2}= \frac{10!}{2}\)minutes

it says "combination", isn't that mean the order of the digits doesn't mater? If the order mater, shouldn't it states "Permutation"?

Bunuel, please can you explain this step (10!/2+10!+10!)∗1/5= 10!/2 minutes?