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Difference between revisions of "2022 AMC 12B Problems/Problem 17"

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<math>\textbf{(A) }144 \qquad \textbf{(B) }240 \qquad \textbf{(C) }336 \qquad \textbf{(D) }576 \qquad \textbf{(E) }624</math>
 
<math>\textbf{(A) }144 \qquad \textbf{(B) }240 \qquad \textbf{(C) }336 \qquad \textbf{(D) }576 \qquad \textbf{(E) }624</math>
  
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==Solution (quick and easy)==
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==Solution 1 (Independence and One-to-One Correspondence)==
  
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Realize the configurations of rows and columns to obtain sums of one of each of <math>1, 2, 3,</math> and <math>4</math> are independent of each other. Then the answer is simply <math>(4!)^2=\boxed{\textbf{(D) 576}}</math>.
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Note that the configurations of row sums and column sums are independent of each other. In addition, the arrays and the sum configuration have one-to-one correspondence. Therefore, the answer is simply <math>(4!)^2=\boxed{\textbf{(D) }576}.</math>
  
 
~bad_at_mathcounts
 
~bad_at_mathcounts
  
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==Solution (Linear transformation, permutation)==
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==Solution 2 (Linear Transformation and Permutation)==
  
 
In this problem, we call a matrix that satisfies all constraints given in the problem a feasible matrix.
 
In this problem, we call a matrix that satisfies all constraints given in the problem a feasible matrix.
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Second, we observe that there is a unique benchmark matrix, as shown below:
 
Second, we observe that there is a unique benchmark matrix, as shown below:
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<cmath>
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<cmath>\left[
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\[
 
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\left[
 
 
   \begin{array}{cccc}
 
   \begin{array}{cccc}
 
     0 & 0 & 0 & 1 \\
 
     0 & 0 & 0 & 1 \\
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     1 & 1 & 1 & 1 \\
 
     1 & 1 & 1 & 1 \\
 
   \end{array}
 
   \end{array}
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\right]
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\right]</cmath>
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\]
 
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</cmath>
 
−
 
 
 
With above observations, we now count the number of feasible matrixes.
 
With above observations, we now count the number of feasible matrixes.
 
We construct a feasible matrix in the following steps.
 
We construct a feasible matrix in the following steps.
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Following from the rule of product, the total number of feasible matrixes is <math>4! \cdot 4! =
 
Following from the rule of product, the total number of feasible matrixes is <math>4! \cdot 4! =
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\boxed{\textbf{(D) 576}}</math>.
+
\boxed{\textbf{(D) }576}</math>.
  
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
  
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==Alternate Solution (From outside to inside)==
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==Solution 3 (Multiplication Principle)==
 +
 
 +
 
 +
==Solution 4 (Multiplication Principle)==
 
Since exactly <math>1</math> row sum is <math>4</math> and exactly <math>1</math> column sum is <math>4</math>, there is a unique entry in the array such that it, and every other entry in the same row or column, is a <math>1.</math> Since there are <math>16</math> total entries in the array, there are <math>16</math> ways to choose the entry with only <math>1</math>s in its row and column.
 
Since exactly <math>1</math> row sum is <math>4</math> and exactly <math>1</math> column sum is <math>4</math>, there is a unique entry in the array such that it, and every other entry in the same row or column, is a <math>1.</math> Since there are <math>16</math> total entries in the array, there are <math>16</math> ways to choose the entry with only <math>1</math>s in its row and column.
  
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WLOG, let that entry be in the bottom right corner of the square. Then, the remaining empty squares are the <math>4</math> center squares. Of these, one of the columns of the empty <math>2\textrm{x}2</math> array will have <math>2</math> <math>1</math>s and the other column will have <math>1</math> <math>1.</math> That happens if and only if exactly <math>1</math> of the remaining squares is filled with a <math>0</math>, and there are <math>4</math> ways to choose that square. Filling that square with a <math>0</math> and the other <math>3</math> squares with <math>1</math>s completes the grid.
 
WLOG, let that entry be in the bottom right corner of the square. Then, the remaining empty squares are the <math>4</math> center squares. Of these, one of the columns of the empty <math>2\textrm{x}2</math> array will have <math>2</math> <math>1</math>s and the other column will have <math>1</math> <math>1.</math> That happens if and only if exactly <math>1</math> of the remaining squares is filled with a <math>0</math>, and there are <math>4</math> ways to choose that square. Filling that square with a <math>0</math> and the other <math>3</math> squares with <math>1</math>s completes the grid.
  
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All in all, there are <math>4*9*16=\boxed{576}</math> ways to complete the grid.
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All in all, there are <math>4\cdot9\cdot16=\boxed{\textbf{(D) }576}</math> ways to complete the grid.
 +
 
 +
~pianoboy
  
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pianoboy.
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~mathboy100 (minor <math>\LaTeX</math> fix)
−
~mathboy100 (minor LaTeX fix)
 
  
 
==Video Solution==
 
==Video Solution==

Revision as of 23:30, 9 January 2023

Problem

How many $4 \times 4$ arrays whose entries are $0$s and $1$s are there such that the row sums (the sum of the entries in each row) are $1, 2, 3,$ and $4,$ in some order, and the column sums (the sum of the entries in each column) are also $1, 2, 3,$ and $4,$ in some order? For example, the array \[\left[ \begin{array}{cccc} 1 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 1 & 1 & 1 & 1 \\ 0 & 1 & 0 & 0 \\ \end{array} \right]\] satisfies the condition.

$\textbf{(A) }144 \qquad \textbf{(B) }240 \qquad \textbf{(C) }336 \qquad \textbf{(D) }576 \qquad \textbf{(E) }624$

Solution 1 (Independence and One-to-One Correspondence)

Note that the configurations of row sums and column sums are independent of each other. In addition, the arrays and the sum configuration have one-to-one correspondence. Therefore, the answer is simply $(4!)^2=\boxed{\textbf{(D) }576}.$

~bad_at_mathcounts

Solution 2 (Linear Transformation and Permutation)

In this problem, we call a matrix that satisfies all constraints given in the problem a feasible matrix.

First, we observe that if a matrix is feasible, and we swap two rows or two columns to get a new matrix, then this new matrix is still feasible.

Therefore, any feasible matrix can be obtained through a sequence of such swapping operations from a feasible matrix where for all $i \in \left\{ 1, 2, 3 ,4 \right\}$, the sum of entries in row $i$ is $i$ and the sum of entries in column $i$ is $i$, hereafter called as a benchmark matrix.

Second, we observe that there is a unique benchmark matrix, as shown below: \[\left[ \begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ \end{array} \right]\] With above observations, we now count the number of feasible matrixes. We construct a feasible matrix in the following steps.

Step 1: We make a permutation of rows of the benchmark matrix.

The number of ways is $4!$.

Step 2: We make a permutation of columns of the matrix obtained after Step 1.

The number of ways is $4!$.

Following from the rule of product, the total number of feasible matrixes is $4! \cdot 4! = \boxed{\textbf{(D) }576}$.

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Solution 3 (Multiplication Principle)

Solution 4 (Multiplication Principle)

Since exactly $1$ row sum is $4$ and exactly $1$ column sum is $4$, there is a unique entry in the array such that it, and every other entry in the same row or column, is a $1.$ Since there are $16$ total entries in the array, there are $16$ ways to choose the entry with only $1$s in its row and column.

WLOG, let that entry be in the top-left corner of the square. Note that there is already $1$ entry numbered $1$ in each unfilled row, and $1$ entry numbered $1$ in each unfilled column. Since exactly $1$ row sum is $1$ and exactly $1$ column sum is $1$, there is a unique entry in the $3\textrm{x}3$ array of the empty squares such that it, and every other entry in the same row or column in the $3\textrm{x}3$ array is a $0.$ Using a process similar to what we used in the first paragraph, we can see that there are $9$ ways to choose the entry with only $0$ in its row and column (in the $3\textrm{x}3$ array).

WLOG, let that entry be in the bottom right corner of the square. Then, the remaining empty squares are the $4$ center squares. Of these, one of the columns of the empty $2\textrm{x}2$ array will have $2$ $1$s and the other column will have $1$ $1.$ That happens if and only if exactly $1$ of the remaining squares is filled with a $0$, and there are $4$ ways to choose that square. Filling that square with a $0$ and the other $3$ squares with $1$s completes the grid.

All in all, there are $4\cdot9\cdot16=\boxed{\textbf{(D) }576}$ ways to complete the grid.

~pianoboy

~mathboy100 (minor $\LaTeX$ fix)

Video Solution

https://youtu.be/_dN_ZHiaiko

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

See Also

2022 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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 12 Problems and Solutions

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