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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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!

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

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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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?

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 \(= (10_P_8 + 10_P_9 + 10_P_{10})*12\)

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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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!!!!

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
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