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I'm facing difficulties while solving this problem: A password contains 5 digit numbers and 5 small english letters. how many possible passwords can be made?

If I solve it like this: 26^5 x 10^5 this will not count passwords that are mixed and digit-first passwords.

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    $\begingroup$ You're right that $26^5 10^5$ is the number of passwords that match the pattern LLLLLNNNNN. How many patterns are possible? $\endgroup$
    – Karl
    Jan 1 at 23:51
  • $\begingroup$ @Karl Each password must have 5 digits and 5 letters. Other than this there are no more restrictions. So any pattern that has 5 digits and 5 letters, is allowed. $\endgroup$
    – Alan Ari
    Jan 2 at 0:00
  • $\begingroup$ How many ways are there of choosing $~5~$ positions out of $~10,~$ sampling without replacement, where order of selection is regarded as unimportant? The $~5~$ selected positions can be regarded as the positions that the digits will occupy, with the letters going in the remaining positions. $\endgroup$ Jan 2 at 0:04
  • $\begingroup$ @user2661923 Is this a question or what? In mine the order is important and it is with replacement/repetition $\endgroup$
    – Alan Ari
    Jan 2 at 0:09
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    $\begingroup$ I don't know what else to say. Is this an assigned problem from a math course? Perhaps you could ask your teacher for help, or ask the school if a tutor is available? $\endgroup$ Jan 2 at 0:36

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Using the work you have already done you have calculated the probability of a password with the position of the numbers and digit fixed. Namely the first five are letters only and the next five are digits only, but this is only one possible way of fixing the positions of the letters and digits. How many ways can they be fixed? Would it be possible to calculate the probability for each of the many ways of fixing the positions? How would you do so?

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  • $\begingroup$ Oh, then multiplying my work by 10 will be the answer? $\endgroup$
    – Alan Ari
    Jan 2 at 0:56
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    $\begingroup$ are there only 10 ways to fix the positions? $\endgroup$ Jan 2 at 0:57
  • $\begingroup$ NNNNNLLLLL will be 26^5 x 10^5. LLLLLNNNNN will be also 26^5 x 10^5. Continuing like this it will have 2^10 different permutation. So the answer will be: 2^10 x 26^5 × 10^5 right? $\endgroup$
    – Alan Ari
    Jan 2 at 1:04
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    $\begingroup$ not exactly $2^{10}$ since everytime a digit or letter is used the chances of the next draw get affected it is not $50|50$ anymore. This means the positions of letters and digits are being chosen "without replacement". $\endgroup$ Jan 2 at 1:14
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    $\begingroup$ Lets put an $L$ in every position that's fixed for a letter. You are trying to figure out how many different positions the five $L$s can take out of $10$ possible positions. $\endgroup$ Jan 2 at 1:58

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