A farmer's rectangular field is partitioned into $2$ by $2$ grid of $4$ rectangular sections as shown in the figure. In each section the farmer will plant one crop: corn, wheat, soybeans, or potatoes. The farmer does not want to grow corn and wheat in any two sections that share a border, and the farmer does not want to grow soybeans and potatoes in any two sections that share a border. Given these restrictions, in how many ways can the farmer choose crops to plant in each of the four sections of the field?
[asy] draw((0,0)--(100,0)--(100,50)--(0,50)--cycle); draw((50,0)--(50,50)); draw((0,25)--(100,25)); [/asy]
$\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 64 \qquad \textbf{(C)}\ 84 \qquad \textbf{(D)}\ 90 \qquad \textbf{(E)}\ 144$
There are $4$ possibilities for the top-left section. It follows that the top-right and bottom-left sections each have $3$ possibilities, so they have $3^2=9$ combinations. We have two cases:


The top-right and bottom-left sections have the same crop.
Note that $3$ of the $9$ combinations of the top-right and bottom-left sections satisfy this case, from which the bottom-right section has $3$ possibilities. Therefore, there are $4\cdot3\cdot3=36$ ways in this case. 
The top-right and bottom-left sections have different crops.
Note that $6$ of the $9$ combinations of the top-right and bottom-left sections satisfy this case, from which the bottom-right section has $2$ possibilities. Therefore, there are $4\cdot6\cdot2=48$ ways in this case. 

Together, the answer is $36+48=\boxed{\textbf{(C)}\ 84}.$
~Arcticturn ~MRENTHUSIASM