Was just wondering whether anybody could help me with the following question:
A system accepts only passwords that contain:
• between 6 and 8 symbols: digit (0-9) or letter (a-z,A-Z);
• at least 2 letters and 2 digits; How many passwords are accepted by this system?
My solution would be to do the following:
If there are 6 symbols then we can have:
$\ \ \ $• 2 letters 4 digits – (2+4)!/2!4!=15. 15 configurations x(52^2 x 10^4) possibilities
$\ \ \ $• 3 letters 3 digits – (3+3)!/3!3!=20. 20 configurations x(52^3 X 10^3)possibilities
$\ \ \ $• 4 letters 2 digits – (4+2)!/4!2!=15. 15 configurations x(52^4 X 10^2) possibilities
If there are 7 symbols we can have:
$\ \ \ $• 2 letters 5 digits – 7!/2!5! = 21. 21 configurations x (52^2x 10^5) possibilities
$\ \ \ $• 3 letters 4 digits – 7!/3!4! = 35. 35 configurations x (52^3x10^4) possibilities
$\ \ \ $• 4 letters 3 digits – 7!/4!3! = 35 35 configurations x (52^4x10^3) possibilities
$\ \ \ $• 5 letters 2 digits – 7!/5!2! = 21 21 configurations x (52^5x10^2) possibilities
If there are 8 symbols we can have:
$\ \ \ $• 2 letters 6 digits – 8!/2!6!= 28. 28 configurations x(52^2 x 10^6) possibilities
$\ \ \ $• 3 letters 5 digits – 8!/3!5!=56. 56 configurations x(52^3x10^5) possibilities
$\ \ \ $• 4 letters 4 digits – 8!/4!4!=70. 70 configurations x(52^4x 10^4) possibilities
$\ \ \ $• 5 letters 3 digits - 8!/5!3!=56. 56 configurations x(52^5x10^3) possibilities
$\ \ \ $• 6 letters 2 digits – 8!/6!2!=28. 28 configurations x(52^6 x10^2) possibilities.
The answer I obtained was the addition of all of these values.
Can anyone help?
Thanks in advance!