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

Revision as of 00:06, 13 June 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)

Solution 5 (Meta Solving/Conjectures)

Note that swapping any two rows or any two columns or both from the given example array, leads to a new array that satisfies the condition. There are $4$ rows, and you choose $2$ to swap, so $4$ ! / ( $2$ ! $2$ !) = $6$ ; likewise for columns. Therefore, $6$ * $6$ = $36$ . Clearly, the final answer must be divisible by $36$ , so we can eliminate B, C and E.

Now, you are in exam mode with not much time left. You see that $144$ = $4$ * $36$ while $576$ = $16$ * $36$ . There are $16$ elements in the array ( $4$ rows and $4$ columns), so you go with $\boxed{\textbf{(D) }576}$ , and this is indeed the correct answer!

- alphamugamma

Video Solution

https://youtu.be/_dN_ZHiaiko

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

Video Solution

https://youtu.be/JVDlHCSPF6k

@@ Line 82: / Line 82: @@
 ==Solution 5 (Meta Solving/Conjectures)==
-Note that swapping any two rows or any two columns or both from the given example array, leads to a new array that satisfies the condition. There are <math>4</math> rows, and you choose <math>2</math> to swap, so <math>4</math>! / <math>2</math>! <math>2</math>! = <math>6</math>; likewise for columns. Therefore, <math>6</math> * <math>6</math> = <math>36</math>. Clearly, the final answer must be divisible by <math>36</math>, so we can eliminate B, C and E.
+Note that swapping any two rows or any two columns or both from the given example array, leads to a new array that satisfies the condition. There are <math>4</math> rows, and you choose <math>2</math> to swap, so <math>4</math>! / (<math>2</math>! <math>2</math>!) = <math>6</math>; likewise for columns. Therefore, <math>6</math> * <math>6</math> = <math>36</math>. Clearly, the final answer must be divisible by <math>36</math>, so we can eliminate B, C and E.
-Now, you are in exam mode with not much time left. You see that <math>144</math> = <math>4</math> * <math>36</math> while <math>576</math> = <math>16</math> * <math>36</math>. There are <math>16</math> elements in the array (<math>4</math> rows and <math>4</math> columns), so you go with D, <math>576</math>, and this is indeed the correct answer!
+Now, you are in exam mode with not much time left. You see that <math>144</math> = <math>4</math> * <math>36</math> while <math>576</math> = <math>16</math> * <math>36</math>. There are <math>16</math> elements in the array (<math>4</math> rows and <math>4</math> columns), so you go with <math>\boxed{\textbf{(D) }576}</math>, and this is indeed the correct answer!
 - alphamugamma

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

Page

Toolbox

Search

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

Revision as of 00:06, 13 June 2023

Contents

Problem

Solution 1 (One-to-One Correspondence)

Solution 2 (Linear Transformation and Permutation)

Solution 3 (Multiplication Principle)

Solution 4 (Multiplication Principle)

Solution 5 (Meta Solving/Conjectures)

Video Solution

Video Solution

See Also