Olympiad Maths Prep

Track / Stage 3 / 134 of 260 #134 of 2000

Problem 134

AMC 10/12, early questions
Combinatorics Difficulty 3.4 Find the answer

A farmer's rectangular field is partitioned into 22 by 22 grid of 44 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]
(A) 12(B) 64(C) 84(D) 90(E) 144\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 64 \qquad \textbf{(C)}\ 84 \qquad \textbf{(D)}\ 90 \qquad \textbf{(E)}\ 144

Official solution

There are 44 possibilities for the top-left section. It follows that the top-right and bottom-left sections each have 33 possibilities, so they have 32=93^2=9 combinations. We have two cases:

The top-right and bottom-left sections have the same crop.
Note that 33 of the 99 combinations of the top-right and bottom-left sections satisfy this case, from which the bottom-right section has 33 possibilities. Therefore, there are 433=364\cdot3\cdot3=36 ways in this case.
The top-right and bottom-left sections have different crops.
Note that 66 of the 99 combinations of the top-right and bottom-left sections satisfy this case, from which the bottom-right section has 22 possibilities. Therefore, there are 462=484\cdot6\cdot2=48 ways in this case.

Together, the answer is 36+48=(C) 84.36+48=\boxed{\textbf{(C)}\ 84}.
~Arcticturn ~MRENTHUSIASM

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.