Random Function size I am new to cryptography. I have question regarding the size of the set of random functions. I found the text from "Introduction to modern cryptography" by Katz and Lindell, in another thread which also relates to my question: Difference between Pseudorandom Function vs randomly chosen function

The text from "Introduction to modern cryptography" by Katz and Lindell (chapter 3.5.1) from the thread:


  Since the notion of choosing a function at random is less familiar
  than the notion of choosing a string at random, it is worth spending a
  bit more time on this idea. From a mathematical point of view, we can
  consider the set $\text{Func}_n$  of all functions mapping $n$ -bit strings to $n$
  -bit strings; this set is finite (as we will see in a moment), and so randomly selecting a function mapping $n$ -bit strings to $n$ -bit strings
  corresponds exactly to choosing an element uniformly at random from
  this set. 
  
  How large is the set $\text{Func}_n$ ? A function $f$  is exactly specified by
  its value on each point in its domain; in fact, we can view any
  function (over a finite domain) as a large look-up table that stores
  $f(x)$  in the row of the table labeled by $x$ . For $f \in\text{Func}_n$ , the
  look-up table for $f$   has $2^n$   rows (one for each point of the
  domain $\{0,1\}^n$  ) and each row contains an $n$ -bit string (since the
  range of $f$   is ${0,1}^n$  ). Any such table can thus be represented
  using exactly $n\cdot 2^n$   bits. 
  
  Moreover, the functions in $\text{Func}_n$  are in one-to-one correspondence
  with look-up tables of this form; meaning that they are in one-to-one
  correspondence with all strings of length $n\cdot 2^n$. 
  
  We conclude that the size of $\text{Func}_n$  is $2^{(n\cdot 2^n)}$. ...


Assume $n = 2$

Than we have following mapping  

00 00  
01 01  
10 10  
11 11  

00 11  
01 00  
10 01  
11 10  

00 10  
01 11  
10 00  
11 01  

00 01  
01 10  
10 11  
11 00 


if we rearrange so that each each row of $f(x)$ corresponds to each $x$. i.e. concatenating we will have following rows.  

00 01 10 11  
11 00 01 10  
10 11 00 01  
01 10 11 00 


Each row will contain $n\cdot2^n$ bits. i.e. our case $2\cdot2^2 = 8$.
And according to my estimate(perhaps incorrect) total number of bits will be $2^n\cdot n\cdot2^n = n\cdot2^{2\cdot n} = 32$. 

While text states the size is $2^{n\cdot2^n}$ i.e. $256$?
Could you please help me to find what I missed in my assumptions and where is my mistake?
Yo are taking only the one-to-one functions, while there are a lot of other functions, to name a few:

00 00
01 00  
10 00  
11 00

00 01
01 01  
10 01  
11 01

00 00
01 00  
10 01  
11 01


as you already mentioned, each table can be represented with $n*2^n$ bits, i.e. strings from the set: $\{0,1\}^{n*2^n}$, each string from this set defines one function from $Func_n$, so the number of such functions is the size of the above set $=2^{n*2^n}$