Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2019 Mock AMC 10B Problems/Problem 21"

(Created page with "==Solution 1: Author:(Shiva Kannan)== There are <math>4! = 24</math> ways to place one of each <math>1, 2, 3,</math> and <math>4</math> in the top row. For the second row, we...")
 
(→‎Solution 1: Author:(Shiva Kannan))
Line 1: Line 1:
 
==Solution 1: Author:(Shiva Kannan)==
 
==Solution 1: Author:(Shiva Kannan)==
−
There are <math>4! = 24</math> ways to place one of each <math>1, 2, 3,</math> and <math>4</math> in the top row. For the second row, we must count all the possible arrangements other than the arrangements in which the value in the first row and a particular column and the value in the second row and that same column, never are the same. The total number of ways regardless of the condition is <math>24</math>. We can use PIE. <math>Number of arrangements in which there is a </math>1<math> under the </math>1<math> in the first row is </math>3! = 6<math>, as the remaining three numbers of the row can be arranged in </math>6<math> ways. By symmetry, this is the same for all </math>4<math> numbers, so the total number of cases in which one value is below itself is </math>24<math>. But we have to correct for overcounting. Number of arrangements in which both </math>1<math> and </math>2<math> are below themselves is </math>2! = 2<math>, as there are </math>2<math> ways to place the remaining two elements. Like this, we can take any pair of numbers from the four given numbers, so the total number of cases in which </math>2<math> values are below themselves is </math>2* {4 \choose 2} = 12<math>. For </math>3<math> numbers being under themselves, we can place the remaining number in </math>1<math> way, and there are three possible sets of </math>3<math> numbers from the four to pick, so the number of cases in which </math>3<math> numbers are below themselves is </math>4<math>. For all four numbers being under themselves, there is only </math>1<math> case. </math>24 - 24 + 12 - 4 + 1 = 9<math>, which yields </math>9$ ways to construct the second row.
+
There are <math>4! = 24</math> ways to place one of each <math>1, 2, 3,</math> and <math>4</math> in the top row. For the second row, we must count all the possible arrangements other than the arrangements in which the value in the first row and a particular column and the value in the second row and that same column, never are the same. The total number of ways regardless of the condition is <math>24</math>. We can use PIE. Number of arrangements in which there is a <math>1</math> under the <math>1</math> in the first row is <math>3! = 6</math>, as the remaining three numbers of the row can be arranged in <math>6</math> ways. By symmetry, this is the same for all <math>4</math> numbers, so the total number of cases in which one value is below itself is <math>24</math>. But we have to correct for overcounting. Number of arrangements in which both <math>1</math> and <math>2</math> are below themselves is <math>2! = 2</math>, as there are <math>2</math> ways to place the remaining two elements. Like this, we can take any pair of numbers from the four given numbers, so the total number of cases in which <math>2</math> values are below themselves is <math>2* {4 \choose 2} = 12</math>. For <math>3</math> numbers being under themselves, we can place the remaining number in <math>1</math> way, and there are three possible sets of <math>3</math> numbers from the four to pick, so the number of cases in which <math>3</math> numbers are below themselves is <math>4</math>. For all four numbers being under themselves, there is only <math>1</math> case. <math>24 - 24 + 12 - 4 + 1 = 9</math>, which yields <math>9</math> ways to construct the second row.

Revision as of 16:09, 4 February 2023

Solution 1: Author:(Shiva Kannan)

There are $4! = 24$ ways to place one of each $1, 2, 3,$ and $4$ in the top row. For the second row, we must count all the possible arrangements other than the arrangements in which the value in the first row and a particular column and the value in the second row and that same column, never are the same. The total number of ways regardless of the condition is $24$. We can use PIE. Number of arrangements in which there is a $1$ under the $1$ in the first row is $3! = 6$, as the remaining three numbers of the row can be arranged in $6$ ways. By symmetry, this is the same for all $4$ numbers, so the total number of cases in which one value is below itself is $24$. But we have to correct for overcounting. Number of arrangements in which both $1$ and $2$ are below themselves is $2! = 2$, as there are $2$ ways to place the remaining two elements. Like this, we can take any pair of numbers from the four given numbers, so the total number of cases in which $2$ values are below themselves is $2* {4 \choose 2} = 12$. For $3$ numbers being under themselves, we can place the remaining number in $1$ way, and there are three possible sets of $3$ numbers from the four to pick, so the number of cases in which $3$ numbers are below themselves is $4$. For all four numbers being under themselves, there is only $1$ case. $24 - 24 + 12 - 4 + 1 = 9$, which yields $9$ ways to construct the second row.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2019_Mock_AMC_10B_Problems/Problem_21&oldid=188609"