= Solution
Write $a=[\mathrm A]$, $b=[\mathrm B]$ and $c=[\mathrm C]$ for the three <concentrations>. The letters denote <chemical species>; the forward arrow represents association of one <molecule> of each reactant into one complex, and the reverse arrow represents dissociation. The positive constants $k_+$ and $k_-$ are the respective <reaction rate constants>. For an elementary <chemical reaction>, the <law of mass action> gives forward and reverse <reaction rates>
$$
R_+=k_+ab,\qquad R_-=k_-c.
$$
Every forward event removes one <molecule> of each reactant and adds one complex; a reverse event reverses those changes. Thus the <reaction-rate equations> are
$$
\boxed{\dot a=-k_+ab+k_-c,\qquad
\dot b=-k_+ab+k_-c,\qquad
\dot c=k_+ab-k_-c.}
$$
The units of $k_+$ are inverse <concentration> per unit time, whereas $k_-$ has units of inverse time. Adding the equations gives the independent <conservation laws>
$$
\boxed{a+c=a(0)+c(0),\qquad b+c=b(0)+c(0).}
$$
In particular $a-b$ is constant, but this is a consequence of those two <conservation laws>. The total $a+b+c$ need not be constant: association changes the number of separate <molecules>.
For the general <stoichiometry>, take $p,q$ to be positive <integers>. The net <reaction rate> and <stoichiometric vector> are
$$
R=k_+a^pb^q-k_-c,\qquad \boldsymbol\nu=(-p,-q,1)^{\mathsf T}.
$$
Consequently
$$
\boxed{\dot a=-pR,\qquad \dot b=-qR,\qquad \dot c=R.}
$$
The factors $p,q$ outside $R$ count <molecules> consumed per event; the powers inside $R$ express the <law of mass action>. The corresponding <conservation laws> are $a+pc=\text{constant}$ and $b+qc=\text{constant}$. No <factorial> is required in this deterministic convention for the <reaction rate constant>.
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