MATH SOLVE

4 months ago

Q:
# Your company requires user passwords that can be made up of combinations of 21 (no caps) letters and 10 numerals only, in any order. Each valid password must be a string of these no more than 18 in length, but atleast 14 in length.(a) How many different passwords are there?(b) Suppose now that each password must contain at least 15 numerals. Now how many possible passwords are there now?

Accepted Solution

A:

Answer:a)There are [tex]7,2236*10^{26}[/tex] possible different passwords.b)There are [tex]3,0784*10^{19}[/tex] possible different passwords.Step-by-step explanation:a) Each character of the string could be any of the 10 numerals or 21 letters, that means, each character may be any of these 31 different options.If the password had only one character, there would be 31 different possible passwords, if it were a two characters string, there would be [tex]31^{2}=961[/tex], because in the first place we can place any of the 31 characters, and then in the second place, we can do the same, we are asked of combinations without taking into account order of the characters.A generalized form, to know how many different passwords we can make of n characters is [tex]31^{n}[/tex]. If we have more than one possible lengths, we add those cases, in our case, we have 14 thru 18 characters, all that could be filled by any of the 31 possible characters.[tex]\[\sum_{i=14}^{18}31^{i} =31^{14}+31^{15} +31^{16}+31^{17}+31^{18}=7,2236*10^{26} Β Β \\\][/tex].b) If each password must contain at least 15 numerals, that means the minimum length of a password is 15 now. We proceed the same as the a) point, but taking into account that for 15 characters, there is not 31 possible characters anymore, but 10.Doing each individual length:[tex]Passwords_{length=15} =10^{15}[/tex]That is because we would multiply 10 possible options by the next possible 10 options and so on 15 times.[tex]Passwords_{length=16} =10^{15}*31[/tex]Now, for the sixteenth character, we multiply by the now 31 possible options.[tex]Passwords_{length=17} =10^{15}*31^{2} \\Passwords_{length=18} =10^{15}*31^{3}[/tex]For lengths 17 and 18, we do similar stuff, multiplying by 31 for each new character.To know all the possible combinations, we add the combinations for each length:[tex]\sum_{i=15}^{18}Passwords_{lenght=i}=3,0784*10^{19}[/tex]We can see there is approximately 7 orders of magnitude less with the restriction than there is without it.