A password to a certain database consists of digits that can

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.

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.

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.

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!

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

From the digits (10 possibilities), a password of at least 8 different digits is constructed.

Possibilities:

- 8-digit password
- 9 digit password
-10 digit password

For the 8-digit password:
The total array is:
8P10 =10!/(10-8)!

For the 9-digit password:
The total array is:
9P10=(10!)/(10-9)!

For the 10-digit password:
The settlement total is:
10P10=(10!)/(10-10)!

That is total passwords:

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

Keep in mind: 0! = 1

Let's continue:

10!(1/2 + 2)

10!(5/2)

As each password generation takes 12 seconds, and we are asked the total time to generate all passwords in minutes.

Keep in mind 12 seconds = 12/60 minutes = 1/5 minutes.

Then:

10!x(5/2)x(1/5)

10!x(1/2)

Answer D

Since it has only been mentioned that a password is an eight-digit number.
The first digit can also take the digit for its beginning value. Since repetition is not allowed.
Going by the order from left to right :
The number of possibilities is:
If the password is an eight-digit number : 10*9*8*7*6*5*4*3. = \(\frac{10!}{2!}\)
If the password is a nine-digit number: 10*9*8*7*6*5*4*3*2. = 10!
If the password is a ten-digit code:  10*9*8*7*6*5*4*3*2. = 10!
A total of 2*(10!) + \(\frac{10!}{2}\) passwords are possible.
\(\frac{\left(\left(10!\right)\cdot5\right)}{2}\) .
Since in a minute they can complete 5 passwords, In order to assure they type the correct password :
\(\frac{\left(\left(10!\right)\left(\frac{5}{2}\right)\right)}{5}\)
\(\frac{10!}{2}\)

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?

An 8-digit password is different from an 8-digit number in that the password can begin with 0, whereas a number cannot.

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