How much possibilities do we have when generating two words and a special character at a random place? There is a method for generating human-memorable passwords, we just need to prove that it is enough strong. Strong means too much possibilities to quickly brute-force all the possible outputs (or other way, ex.: dictionary attack). 
What do we have? 
We have a public dictionary wordlist of 8000 unique words (one word is at least 9 characters long, ex.: 10, 11, 12, etc., the average length is: 11. Using the 26 lowercase english letters). The words are human-memorable, got them from a spell checker, ex.: https://www.libreoffice.org/download/libreoffice-fresh/ - search for "dictionary". 
We have 27 different special characters. 
The method: 
We select two random words from the wordlist, the selected words can be the same, space is the delimiter, example: 
concretely paintball
doorframe magnetized
uncertainty chelonian
popularizer popularizer
supercritical liquidate

We put a random special character to a random place to this line (the line is the two words and a space between them), example: 
conc>retely paintball
doorframe magneti]zed
$uncertainty chelonian
popularizer) popularizer
supercritical liquidate=

One line is one password. 
When we talk about random, it is ensured, that it is really random. 
The Big question is: what is the formula to know how many different passwords could there be? So: 2 random words + 1 space + 1 random special character at a random place. 
If we know how many different passwords could be generated with this method, we can prove that it is an enough strong method to generate human-memorable passwords or not. 
 A: $8000\times 8000$ for the word choices, of course.
How may ways there are to add a "random special character" depends on how long the words are, but they're probably usually not more than 14 characters each, which gives 30 positions. There are 30 printable non-alphanumerics in ASCII, leading to at most $30\times 30$ ways to put in your special character.
All in all you're sampling from a space of somewhat less than $2^{36}$ possible passwords.
As Xkcd famously points out, four words drawn from a 2000-word list will beat that, and then you don't even need to require that the words are long (and thus more difficult to remember than the simple words you can populate a 2000-word list with).
A: There are $8000 \times 8000$ different ways of taking two words. 
In average each pair is ** 11 letters long** and there are two words. There are  ** 23 characters** from the dictionary (including the space). That means we have  24 places to put the special character (We can put it behind each character and also in first position).
That makes an approximation of $24 \times 64 · 10^6$ different passwords for each special character. As there are 27 special characters that makes it  $27 \times 24 \times 64 ·10^6$ passwords.
Remember this is an approximation based on the average length hypothesis
