If the rate law order is three, does that mean that three bodies collide at some elementary step?

I am doing a chemistry review about rates, and I get asked "Which of these statements is false?"


The answer is 



> 
> In reactions that are second order in one reactant and first order in another, the slow step generally involves a three-body collision of the reactants.
> 
> 
> 


I imagine it looks like


$$
rate = k[A]^2[B]^1
$$


So at the slowest elementary step there should be
$$
2A + B = something
$$


Which is a three way collision between two A molecules and a B. 


What am I doing wrong?


EDIT:


Question number 2 (<http://www.mesacc.edu/~paudy84101/CHM152F2005/Exam1%20Key.pdf>)


This is not my review, in case you are wondering. I just have the same question.




Let's disprove that statement by counterexample. Here, the slow step is a collision of some intermediate and another thing, but the maths work out that the overall reaction order is third. 


Let's have a two step reaction with the stoichiometry:


$$\ce{2A + B -> C}$$


The mechanism looks like this. The first step is fast and reversible and rapidly establishes equilibrium between $\ce{A + B}$ and an intermediate $\ce{I}$.


$$\ce{A + B <=>[k\_1][k\_{-1}] I}$$


The second step is slow, not so reversible and involves the intermediate $\ce{I}$ reacting with the other equivalent of $\ce{A}$.


$$\ce{I + A ->[k\_2] C}$$


Now the rate of the reaction shall be written as the rate of appearance of $\ce{C}$, because it makes our lives a little easier.


$$\mathrm{rate}=\dfrac{d[\ce{C}]}{dt}=k\_2 [\ce{A}][\ce{I}]$$


However, it's messy to have the intermediate in the rate law. Since the first step was fast and in equilibrium, we can get an expression for $\ce{[I]}$ in terms of $[\ce{A}]$ and $[\ce{B}]$.


$$K\_{eq}=\dfrac{k\_1}{k\_{-1}}=\dfrac{[\ce{I}]}{[\ce{A}][\ce{B}]}\\
[\ce{I}]=K\_{eq}[\ce{A}][\ce{B}]$$


A little substituting and we get:


$$\mathrm{rate}=\dfrac{d[\ce{C}]}{dt}=\dfrac{k\_2 k\_1}{k\_{-1}} [\ce{A}]^2 [\ce{B}]=k\_{obs}[\ce{A}]^2 [\ce{B}]$$


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