If I read the sentence


  Find ten apples and oranges.


Do I need to find ten or twenty pieces of fruit?


Question: Rate-limiting in transition state theory
Answer: In transition state theory (TST, goldbook) one of the necessary assumptions is that reactants and products are in equilibrium. In principle this gives us
\begin{align}\ce{
A &amp;&lt;=&gt; [AB]^{\ddagger} -&gt; B\\
B &amp;&lt;=&gt; [AB]^{\ddagger} -&gt;A\\
A &amp;&lt;=&gt; B,
}\end{align}
and in addition to this, also
\begin{align}\ce{
B &amp;&lt;=&gt; [BC]^{\ddagger} -&gt; C &amp; B &amp;&lt;=&gt; [BD]^{\ddagger} -&gt; D\\
C &amp;&lt;=&gt; [BC]^{\ddagger} -&gt; B &amp; D &amp;&lt;=&gt; [BD]^{\ddagger} -&gt; B\\
B &amp;&lt;=&gt; C                    &amp; B &amp;&lt;=&gt; D,
}\end{align}
as well as
\begin{align}\ce{
C &lt;=&gt; [BC]^{\ddagger} -&gt; &amp; B &lt;=&gt; [BD]^{\ddagger} -&gt; D\\
D &lt;=&gt; [BD]^{\ddagger} -&gt; &amp; B &lt;=&gt; [BC]^{\ddagger} -&gt; C\\
C &lt;=&gt; &amp; B &lt;=&gt; D,
}\end{align}
and 
$$\ce{
C &lt;=&gt; A &lt;=&gt; D\\
[AB]^{\ddagger}&lt;=&gt; [BC]^{\ddagger}&lt;=&gt; [BD]^{\ddagger}
}.$$

You can use TST to estimate rate constants.
$$k = \frac{\mathcal{k}_\mathrm{B} T}{\mathrm{h}} \exp\left\{ −\frac{\Delta^\ddagger{}G^\circ}{\mathcal{R} T}\right\}$$

Since all components have to be in equilibrium, you can predict a ratio between them via Boltzmann statistics, hence:
$$\frac{N([BC]^{\ddagger})}{N([BD]^{\ddagger})} =  \exp\left\{ −\frac{\Delta^\ddagger{}G^\circ([BC]^{\ddagger})-\Delta^\ddagger{}G^\circ([BD]^{\ddagger})}{\mathcal{R} T}\right\}=\frac{N(C)}{N(D)} $$

The preceding equilibrium does not need to be considered to determine the ratio of the products. However, it is important to note, that there are many assumptions necessary for TST to work. For example, at higher temperatures, anharmonic corrections have to be considered. More crucial is the assumption, that the transition can be described as a translatory movement, i.e. it is treated with classical mechanics. 

For a system with a preceding equilibrium, it is well possible, that the TST approximation breaks down completely.