Online database servers often generate random passwords which the user must repeat back to the server to confirm that the request is coming from a human user and not a malicious netbot. The systems for two different airlines use variations of this approach:

Pigeon Airlines generates passwords by randomly selecting seven uppercase letters, without using the same letter twice.

Bumblebee Airways randomly selects six uppercase letters, again without duplication, and appends a randomly selected decimal digit.

a. Calculate the number of different passwords that could be generated for each of these two reservation systems.


 For Pigeon Airlines - Number of Passwords that can be generated are- 

 **26 X 25 X 24 X 23 X 22 X 21 X 20 = 3515312000**

 For Bumblee Airlines - Number of Passwords that can be generated are -

 **26 X 25 X 24 X 23 X 22 X 21 X 10 = 1657656000**

b. If a netbot was to try a brute-force attack on one of these reservation systems by trying to guess the password using random trials, which reservation system would be easier to crack?


 Probability to get Pigeon Airlines Correct = 1/3315312000 = 0.00000000030163
 Probability to get Bumble Bee Correct = 1/1657656000 = 0.00000000060326

 So, it would be easier for him to crack BumbleBee Airlines!

Is my answer correct? Thanks for the help!

Edited my answer to make it more understandable!

  • 1
    $\begingroup$ Check the answer for the first one, should the denominator be $20!$? $\endgroup$ – Macavity Nov 6 '13 at 5:47
  • $\begingroup$ isnt that because - There are 7 positions and every position can have 26 letters but no repetition. Therefore, every position can have 26X25X24X23X22X21. And that should be the final answer. I did the 20! in the denominator to make it easy in calculation $\endgroup$ – CuriousBeing Nov 6 '13 at 5:49
  • $\begingroup$ @Macavity I edited my answer. Is this correct? $\endgroup$ – CuriousBeing Nov 6 '13 at 5:55
  • $\begingroup$ No, the first one still needs one more multiplier for the 7th letter which still has 20 possibilities. For the second one, how does 30 make sense as the last multiplier? There are 10 possible digits in the end unless you are using a rather exotic base. $\endgroup$ – JB King Nov 6 '13 at 6:04
  • 1
    $\begingroup$ Looks good now. $\endgroup$ – Macavity Nov 6 '13 at 6:10

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