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

There are [tex]\displaystyle P(26,\, 3) = \frac{26!}{(26 - 3)!} = 26 \times 25 \times 24 = 15,\!600[/tex] ways to arrange three out of [tex]26[/tex] distinct letters (without replacement.)

Similarly, there are [tex]\displaystyle P(10,\, 4) = \frac{10!}{(10 - 4)!} = 10 \times 9 \times 8 \times 7= 5,\!040[/tex] ways to arrange four out of [tex]10[/tex] distinct numbers (also without replacement.)

Therefore, there are [tex]15,\!600[/tex] possibilities for the three-letter section of this password, and [tex]5,\!040[/tex] 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 [tex]5,\!040[/tex] possibilities. However, if any of the first three letters was changed, the result would be another [tex]5,\!040\![/tex] possibilities, all of which are different from the previous [tex]5,\!040\!\![/tex] possibilities. These two three-letter sections along will give [tex]2 \times 5,\!400[/tex] possibilities. Since there are [tex]15,\!600[/tex] three-letter sections like that, there would be [tex]15,\!600 \times 5,\!400 = 78,\!624,\!000[/tex] possible passwords in total. That gives the number of possible passwords that satisfy these requirements.