Does the mixture of a ligand exchange reaction contain all the complexes obtained in the part reactions?
Yes, all those complexes from all those reactions will be present, and some more. Essentially Loong did all the mathematics in his answer to: Why is ligand substitution only partial with copper(II) ions and ammonia?

Let's shorten this a little bit and write the general reaction equation:
\begin{align}
  \ce{[Cu(H2O)_{(7-n)}]^2+ + n NH3 &amp;&lt;=&gt; [Cu(NH3)_n(H2O)_{(6-n)}]^2+ + n H2O};&amp;
  n&amp;\in\{\mathbb{N}; 1,\dots,6\}
\end{align}
Therefore we can write the general equilibrium constant:
\begin{align}
  K_n &amp;= 
    \frac{{\left[ \ce{[Cu(NH3)_n(H2O)_{(6-n)}]^2+} \right]}}
   {{\left[ \ce{[Cu(H2O)_{(7-n)}]^2+} \right]\left[ \ce{NH3} \right]^n}}\\
%  K_\mathrm{B} &amp;=
%    \frac{{\left[ \ce{[Cu(NH3)6]^2+} \right]}}
%    {{\left[ \ce{[Cu(H2O)6]^2+} \right]\left[ \ce{NH3} \right]^6}} 
%  &amp;&amp;= \prod_{n=1}^{6} K_n
\end{align}

Since all reactions are in equilibrium, all possible species will exist, although they might not be quantifiable. The overall equilibrium depends on the concentration of the involved species, especially ammonia.

If $[\ce{NH3}]$ is low, then a complex with smaller $n$ will be more stable, if you go to higher concentrations, larger $n$ will be preferred.

In all of this it should be mentioned that there are also complexes with vacant coordination sites $\square$, like anything in between
$$\ce{[Cu(H2O)5\square_1]^2+},\dots,
  \ce{[Cu(H2O)3(NH3)2\square_1]^2+},\dots,
  \ce{[Cu(NH3)5\square_1]^2+},\\
  \left(\ce{[Cu(H2O)4\square_2]^2+},\dots,\ce{[Cu(NH3)4\square_2]^2+}\right),$$
and possibly others.