Been trying to do this question since 3 days yet not been able to approach please!!! help

If we disable and technically block all easily guessable weak passwords (specific combinations of letters, dates, names, words of any natural language, such as qwerty, Workplace4, 123de1, Jane2002, GoodPass etc.), the strength of passwords can be evaluated by the total number of different possible passwords that can be formed, according to the actual (in force) password rules. In other words - the more different possible passwords can be created, the more secure the password is.

Evaluate, If and on which level John's assumption holds More specifically • determine how many characters shorter passwords or how many percent shorter passwords John's proposal allows to be used .

• If you ignore the first paragraph because it accounts for a tiny fraction of the passwords, how many characters are in each of the two character sets? How many passwords are there of $n$ characters with each set? – Ross Millikan May 26 '20 at 14:39

I'm not sure I 100% understand your question, but here's my idea: assume the vocabulary size is $$m$$. In total there are $$26+26+10+24 = 86$$ symbols from which you can construct a password. For any $$n$$, there are $$a_n = \binom{86}{n}$$ ways to create a new password length $$n$$, each such list size $$a_n$$ is compared to the vocabulary, let's say there are $$b_n = a_n - r_n$$ unique passwords, where $$r_n$$ is the overlap between $$a_n$$ and vocabulary. Currently there are $$K$$ password options and the password length is $$n_0$$.
What you are after is $$n^{\ast} = \texttt{argmin}_{n} b_n > K$$ which has a solution if $$n^{\ast} < n_0$$
• the smallest $n$ such that $b_n > K$. If this $n < n_0$, the suggestion is correct – Alex May 26 '20 at 17:30
• @mathelete38: $n_0 - n^{\ast}$ – Alex May 27 '20 at 7:28