The password for a computer account has to consist of exactly eight

Question

The password for a computer account has to consist of exactly eight characters. Characters can be chosen from any of the following: letter of the alphabet, numerical digits from 0 to 9, a hyphen, or the exclamation mark. Upper-case letters (e.g., A) are considered different from lower-case letters (e.g., a), and characters can be repeated. Given these rules, how many different passwords are possible?

A. 2^9
B. 2^14
C. 2^18
D. 2^40
E. 2^48

A. 2^9
B. 2^14
C. 2^18
D. 2^40
E. 2^48

BigFatBassist · Answer

E

26 letters in the alphabet, 26*2 = 52 letters, since the password is case-sensitive. Accounting for possibilities of the other special characters we have;
26+26+10+1+1 = 64, or 2^6 possibilities per character of the password.
Total number of possible combinations: (2^6)^8 = 2^48

pkk1611 · Answer

can you please explain the difference between without repetition and with repetition in permutation and combination means

davedekoos · Answer

Hi pkk, In permutations and combinations, if repetition is allowed, then each event is independent from one another. For example, if we are picking a 4 digit code from the numbers 0-9 WITH repetition, then for each selection we can pick any number for each digit. The choices made for the first digit will not affect the choice for the second, third and fourth digits. In this case, the number of permutations would be 10^4 . If we're dealing with combinations (where the order doesn't matter) (i.e. the code 1123 is equivalent to the code 3112, etc.), then the number of combinations is \frac{(r+n-1)!}{(r!)(n-1)!} , where r is the number of digits in our code, and n is the amount of different numbers we can choose (0-9). So in this example the number of combinations with replacement is: \frac{(4+10-1)}{(4!)(10-1)!}=\frac{13!}{(4!)(9!)} . If repetition is not allowed, then each choice made restricts the subsequent choices. In our example, if I were to choose a code of 4 digits without repeating any digits, then the total permutations would be 10!/6! = 10*9*8*7. In other words, I can choose any number for the first digit, any of the 9 remaining numbers for the second digit, any of the 8 remaining numbers for the third digit, and any of the 7 remaining numbers for the last digit. If we're talking about combinations (order is not important), then we just have to take into account all the ways a group of 4 numbers can be arranged, which is 4!. So the number of combinations without replacement would be \frac{10!}{(6!)(4!)} . For a nice detailed explanation with more examples, check out https://www.mathsisfun.com/combinatorics ... tions.html Cheers,

BrentGMATPrepNow · Answer

Note: there are 64 characters to choose from: (26 lowercase letters + 26 uppercase letters + 10 digits + 1 hyphen + 1 exclamation mark = 64)

Take the task of creating a password and break it into  stages .

Stage 1 : Select the 1st character
We can complete stage 1 in  64  ways

Stage 2 : Select the 2nd character
We can complete stage 2 in  64  ways

Stage 3 : Select the 3rd character
We can complete stage 3 in  64  ways 
.
.
.
.
 Stage 8 : Select the 8th character
We can complete stage 8 in  64  ways

By the Fundamental Counting Principle (FCP), we can complete all 8 stages (and thus create an 8-character code) in  (64)(64)(64)(64)(64)(64)(64)(64)  ways 
This is the same as 64^8, but this is not one of the answer choices.

So, replace 64 with  2^6  to get: 
64^8 = ( 2^6 )^8
= 2^48
= E
 
Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. For more information about the FCP, watch our free video:  https://www.gmatprepnow.com/module/gmat-counting/video/775

Cheers,
Brent

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The password for a computer account has to consist of exactly eight

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The password for a computer account has to consist of exactly eight

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