Changing the order of the reaction with change in concentration The reaction

$$\ce{2 A + B + C -&gt; D + 2 E}$$

is found to follow the rate law as

$$r = k[\ce{A}][\ce{B}]^2[\ce{C}]^0$$

If the concentration of $\ce{A}$, $\ce{B}$ and $\ce{C}$ increases two times then the rate of reaction becomes eight times higher:

$$r' = k\cdot 2[\ce{A}]\left(2[\ce{B}]\right)^2\left(2[\ce{C}]\right)^0 = 8k[\ce{A}][\ce{B}]^2$$

Note: there is no change in the order of reaction. I want to change the order of this reaction by changing the concentration of reactants. Can I do this?

Because if we double the concentration of $\ce{A}$, $\ce{B}$ or $\ce{C}$, it does not change the order. But if we increase  the concentration of $\ce{A}$ by 4 times, then we can change order by

$$\left(4[\ce{A}]\right)^2 = \left(2[\ce{A}]\right)^{2+2} = \left(2[\ce{A}]\right)^4$$
You would change the reaction speed, but not the order.

The order of the reaction can be changed by reagent concentration, if it is result of multiple reaction steps with different orders. By variation of concentrations you can manipulate the relative reaction speed,
as the order is often ruled by the slowest reaction.

Let suppose the 2 step reaction 
$$\begin{align}
\ce{2A + B &amp;-&gt; C} \\
\ce{A + A &amp;-&gt;[slow] D} \\
\ce{B + D &amp;-&gt;[fast] C} \\
\end{align}$$

For small $c_A$, the reaction speed
$$\frac{dc_C}{dt}=k_{low A}.{c_A}^2$$.

For high $c_A$, the first reaction becomes faster then the second.

$$\frac{dc_C}{dt}=k_{high A}.{c_A}^2.c_B$$.

You can also change the effective order of reaction, if some reactant is in about constant abundance and is not explicitly considered.

E.g. The speed of the reversed esterification
$$\ce{R1COOR2 + H2O -&gt; R1COOH + R2OH}$$ is the reaction kinetics of the 2nd order. But in abundance of water, it is effectively a reaction of the 1st order:
$$\frac{dc_{\ce{R1COOR2}}}{dt}=-k_1.c_{\ce{R1COOR2}}$$.