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

Revision as of 01:40, 11 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 (One-to-One Correspondence)

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

~bad_at_mathcounts ~MRENTHUSIASM

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)

Of the sixteen entries in one such array, there are six $0$ s and ten $1$ s. The rows and the columns each have zero, one, two, and three $0$ s, in some order. Once we decide the positions of the $0$ s, we form one such array.

We perform the following steps:

There are $\binom41=4$ ways to choose the row with three $0$ s. After that, there are $\binom43=4$ ways to choose the positions of the three $0$ s.

There are $\binom31=3$ ways to choose the column with three $0$ s. After that, there are $\binom32=3$ ways to choose the positions of the two additional $0$ s.

There are $\binom41=4$ ways to choose the position of the last $0:$ It is the intersection of the column of one $0$ in Step 1 and the row of one $0$ in Step 2.

Together, the answer is $4\cdot4\cdot3\cdot3\cdot4=\boxed{\textbf{(D) }576}.$

~MRENTHUSIASM

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.

Without loss of generality, 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 $3\times3$ array of the empty squares such that it, and every other entry in the same row or column in the $3\times3$ 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 $3\times3$ array).

Without loss of generality, 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 $2\times2$ array will have two $1$ s and the other column will have one $1.$ 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)

@@ Line 73: / Line 73: @@
 Without loss of generality, let that entry be in the top-left corner of the square. Note that there is already <math>1</math> entry numbered <math>1</math> in each unfilled row, and <math>1</math> entry numbered <math>1</math> in each unfilled column. Since exactly <math>1</math> row sum is <math>1</math> and exactly <math>1</math> column sum is <math>1</math>, there is a unique entry in the <math>3\times3</math> array of the empty squares such that it, and every other entry in the same row or column in the <math>3\times3</math> array is a <math>0.</math> Using a process similar to what we used in the first paragraph, we can see that there are <math>9</math> ways to choose the entry with only <math>0</math> in its row and column (in the <math>3\times3</math> array).
-Without loss of generality, 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\times2</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.
+Without loss of generality, 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\times2</math> array will have two <math>1</math>s and the other column will have one <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.
 All in all, there are <math>4\cdot9\cdot16=\boxed{\textbf{(D) }576}</math> ways to complete the grid.

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