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Difference between revisions of "2021 Fall AMC 10A Problems/Problem 18"

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- Aidensharp
 
- Aidensharp
  
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==Solution 3(Expected Value)==
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==Solution 3==
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We start by saying there are <math>4</math> choices for the first spot. Then, there are <math>3</math> choices for the second and fourth spots. Let's begin with casework on the third box. <math>\frac {1}{3}</math> of the time both the second box and the fourth box is planted with the same crop, so the third box has <math>3</math> ways to plant the crops. We have an expected value of <math>\frac {108}{3} = 36</math> different crop plantings. Next, if boxes two and four are not the same with <math>\frac {2}{3}</math> probability, we have that the third box can only be <math>2</math> cases. We have <math>\frac {144}{3} = 48</math> expected crop plantings. Therefore, we have <math>36 + 48 = \boxed {\textbf{(C)}84}</math>
+
We start by saying there are <math>4</math> choices for the first spot. Then, there are <math>3</math> choices for the second and fourth spots. Let's begin with casework on the third box. <math>\frac {1}{3}</math> of the time both the second box and the fourth box is planted with the same crop, so the third box has <math>3</math> ways to plant the crops. We have <math>\frac {108}{3} = 36</math> different crop plantings. Next, if boxes two and four are not the same with <math>\frac {2}{3}</math> probability, we have that the third box can only be <math>2</math> cases. We have <math>\frac {144}{3} = 48</math> crop plantings for this case. Therefore, we have <math>36 + 48 = \boxed {\textbf{(C)}84}</math>
  
 
~Arcticturn
 
~Arcticturn

Revision as of 23:50, 23 November 2021

Problem

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] $\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 64 \qquad \textbf{(C)}\ 84 \qquad \textbf{(D)}\ 90 \qquad \textbf{(E)}\ 144$

Solution 1

We will do casework on the type of crops in the field.

Case 1: all of a kind. If all four sections have the same type of crop, there are simply $\underline{4}$ ways to choose crops for the sections.

Case 2: 3 of a kind, 1 of another kind. Since the one of another kind must be adjacent to two of the other crops, when choosing the type of crops in this case, we cannot choose soybeans and potatoes, or corn and wheat. Therefore, there are $4 \cdot 3 - 2 \cdot 2 = 8$ choices for the two crops we choose for the section (notice we did not choose by $2$, since the crop we pick first will be the unique one), and $4$ ways to choose which section the unique crop is planted on. This gives us a total of $4 \cdot 8 = \underline{32}$ ways to choose crops for the sections.

Case 3: 2 of a kind, 2 of another kind. We cannot choose corn and wheat, or soybeans and potatoes, once again, because if we do, the two would have to be adjacent in some way, which the problem disallows. So, there are ${4 \choose 2} - 2 = 4$ ways to choose our two crops (notice that we did choose by $2$, since there are two of both crops). There are ${4 \choose 2} = 6$ ways to choose where one of the crops go, so there are $4 \cdot 6 = \underline{24}$ ways to choose crops for the sections.

Case 4: 2 of a kind, 1 of another kind, 1 of another kind. In cases 3 and 2, we excluded the possibility of choosing bad pairs for our crops (i.e. soybeans and potatoes, or corn and wheat). In this case, it is inevitable that we choose a bad pair, because we are choosing $3$ crops this time. The two sections of the same kind must contain the crop that is not part of the bad pair in the trio: for example, if we choose corn, soybeans and potatoes as our three crop types, nor soybeans and potatoes can be the type which occupies two sections in this case; corn must be the one to do so. There are $4$ ways to choose the crop that is not part of the bad pair, and then $1$ way to choose the bad pair, giving us $4 \cdot 1 = 4$ ways to choose the crops. To separate the bad pair of crops, the two of a kind must be diagonally placed. There are $2$ ways to choose where the two of a kind go, and $2$ ways to choose which of the bad pair goes where, giving us $2 \cdot 2 = 4$ ways to choose the positions for the crops. In total, there are $4 \cdot 4 = \underline{16}$ ways to choose crops for the sections.

Case 5: every single crop. Bad pairs must be on the same diagonal, so there are $2$ ways to choose which pair gets which diagonal, and $2 \cdot 2 = 4$ ways to choose which of each pair goes where on the diagonal, giving us $2 \cdot 4 = \underline{8}$ ways to choose crops for the sections.

Adding up all our values, we get our final answer of $4+32+24+16+8 = \boxed{\textbf{(C)}\ 84}$.

~ihatemath123

Solution 2 (Risky, but Quick)

The top right box has $4$ choices and the top left box has $3$ choices. Thus, it is reasonable to assume that the answer is a multiple of $12$. We know that the answer will not be too small or too large, so the answer is $\boxed{\textbf{(C)}\ 84}$

- Aidensharp

Solution 3

We start by saying there are $4$ choices for the first spot. Then, there are $3$ choices for the second and fourth spots. Let's begin with casework on the third box. $\frac {1}{3}$ of the time both the second box and the fourth box is planted with the same crop, so the third box has $3$ ways to plant the crops. We have $\frac {108}{3} = 36$ different crop plantings. Next, if boxes two and four are not the same with $\frac {2}{3}$ probability, we have that the third box can only be $2$ cases. We have $\frac {144}{3} = 48$ crop plantings for this case. Therefore, we have $36 + 48 = \boxed {\textbf{(C)}84}$

~Arcticturn

See Also

2021 Fall AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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