Question

A computer password is required to be 7 characters long. How many passwords are possible if the password requires 3 letter(s) followed by 4 digits (numbers 0-9), where no repetition of any letter or digit is allowed

Answer 1

`78,\!624,\!000`.

Explanation:

Note the requirements:

• Repetition of letter or digit is not allowed.
• The order of the letters and digits matters.

Because of that, permutation would be the most suitable way to count the number of possibilities.

There are
`\displaystyle P(26,\, 3) = (26!)/((26 - 3)!) = 26 * 25 * 24 = 15,\!600` ways to arrange three out of
`26` distinct letters (without replacement.)

Similarly, there are
`\displaystyle P(10,\, 4) = (10!)/((10 - 4)!) = 10 * 9 * 8 * 7= 5,\!040` ways to arrange four out of
`10` distinct numbers (also without replacement.)

Therefore, there are
`15,\!600` possibilities for the three-letter section of this password, and
`5,\!040` possibilities for the four-digit section. What if these two parts are combined?

Consider: if the first three letters of the password were fixed, then there would be
`5,\!040` possibilities. However, if any of the first three letters was changed, the result would be another
`5,\!040\!` possibilities, all of which are different from the previous
`5,\!040\!\!` possibilities. These two three-letter sections along will give
`2 * 5,\!400` possibilities. Since there are
`15,\!600` three-letter sections like that, there would be
`15,\!600 * 5,\!400 = 78,\!624,\!000` possible passwords in total. That gives the number of possible passwords that satisfy these requirements.