= Solution
{c}
Write $n=n_1$ and $m=n_2$. The three <stoichiometric vectors> are $(1,0)$, $(0,1)$ and $(-1,-1)$. The <law of mass action> gives the respective <reaction propensity functions>
$$
a_1(n,m)=2,\qquad a_2(n,m)=\frac n2,\qquad a_3(n,m)=nm.
$$
In particular, producing the second <chemical species> does not consume the first <chemical species>. Since the two reactants in the third <chemical reaction> are different <chemical species>, the number of possible pairs is $nm$.
For each <Markov chain state>, the <chemical master equation> adds incoming <probability> flux and subtracts outgoing <probability> flux:
$$
\boxed{\begin{aligned}
\frac{\partial p(n,m,t)}{\partial t}
={}&2p(n-1,m,t)+\frac n2p(n,m-1,t)\\
&+(n+1)(m+1)p(n+1,m+1,t)\\
&-\left(2+\frac n2+nm\right)p(n,m,t).
\end{aligned}}
$$
Here $p(n,m,t)=0$ if either index is negative, and the <initial condition> is $p(n,m,0)=\mathbf1_{\{n=m=0\}}$. These conventions also give the correct <chemical master equation> on the coordinate axes: for example, the third <chemical reaction> has zero outgoing <reaction propensity function> when either <chemical species> is absent.
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