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A password to a certain database consists of digits that cannot be repeated. If the password is known to consist of at least 8 digits and it takes 12 seconds to try one combination, what is the amount of time, in minutes, necessary to guarantee access to database?
A. 8!/5
B. 8!/2
C. 8!
D. 10!/2
E. 5/2.10!
OA:DThere 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
How can the number of possibilities for the first digit in a 8 digit password be 10? If we choose 0 as the first digit, won't the password end up being 7 digit?