# How many different passwords of length 6 can be formed…

Each user on a computer system has a password, which is six characters long, where each character is an uppercase letter or digit. Each password must contain at least one digit. How many possible passwords are there?

Why am I incorrect in reasoning that the answer is $10*36^5$? If every password must contain a digit, then there are only $10$ ways to choose one character in the password and $36^5$ ways to choose the other 5 characters.

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You’re not taking into account that the digit can occur in any of the six positions. There are $10\cdot36^5$ passwords that begin with a digit. There are also $10\cdot36^5$ passwords that end with a digit; some of these also begin with a digit and have already been counted, but some do not, so your figure of $10\cdot36^5$ is necessarily too small.

The easiest way to count the acceptable passwords is to note that there are $36^6$ six-character strings made up of upper-case letters and digits, and $26^6$ of them are made up entirely of letters, so there are $36^6-26^6$ that include at least one digit.

It is possible to count them directly, but the counting is more complicated. For each of the $6$ positions in the password there are $10\cdot36^5$ passwords having a digit in that position, so to a first approximation there are $6\cdot10\cdot36^5$ acceptable passwords. However, as noted in the first paragraph, this counts some passwords more than once. For each pair of positions in the password there are $10^2\cdot36^4$ passwords having digits in both of those positions, and all of these passwords have been counted twice. Since there are $\binom62$ pairs of positions, we must subtract $\binom62\cdot10^2\cdot36^4$ to get rid of the double-counting. Unfortunately, this overcompensates, and there are further corrections to be made. The net result, given by the inclusion-exclusion principle, is

$$\sum_{k=1}^6(-1)^{k+1}\binom6k10^k36^{6-k}\;.$$

Either way, the result is $1,867,866,560$.

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Why does $36^6-26^6$ represent ones that include at least one digit? I'm sorry, it is just confusing to me. –  user1038665 Oct 27 '12 at 10:16
@user1038665: There are $36^6$ altogether. There are $26^6$ that have no digit. $36^6-26^6$ is what’s left after you throw away the ones that have no digit, and those are precisely the ones that have at least one digit. –  Brian M. Scott Oct 27 '12 at 10:21
Answers: $36^6$, $26^6$, $36^6-26^6$.