Deriving rate equation of a Chemical reaction

Question

I want to derive the rate equation of $x$, concentration of $X$, in the following reaction: $$ aX + bY \rightarrow Z.$$ I know the answer is $$\frac{dx}{dt}=-kx^ay^b$$ for a constant $k$, but I need more rigorous justification about it.
First, I assumed that the entire volume of the solution is fixed during the reaction.
Then by assuming 'the reaction occurs when $a$ molecules of $X$ (chosen from $x(t)V$ molecules) and two molecules of $Y$ (chosen from $y(t)V$ molecules) meet by $k$ chance', we have: $$\begin{align*}x(t+dt)V-x(t)V&=-k{x(t)V \choose a}{y(t)V \choose b}dt\\&\approx -kx(t)^a y(t)^bV^{a+b}dt \end{align*}$$ By dividing both sides by $Vdt$ and taking $\lim\limits_{dt \rightarrow 0}$, we get $$\frac{dx}{dt}=-kx^a y^b V^{a+b-1}$$ But I have no idea how to remove $V^{a+b-1}$ from the equation.
Could you help me?

Answer 1

Since $V^{a+b-1}$ is a constant, you can let it's product with the constant $k$ be the new $k$. $$\frac{dx}{dt}=kx^ay^bV^{a+b-1}\to kx^ay^b$$ for abitrary constant $k$.