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 "2018 AMC 8 Problems/Problem 16"

(→‎Problem)
m (→‎=Solution)
 
(4 intermediate revisions by 3 users not shown)
Line 4: Line 4:
 
<math>\textbf{(A) }1440\qquad\textbf{(B) }2880\qquad\textbf{(C) }5760\qquad\textbf{(D) }182,440\qquad \textbf{(E) }362,880</math>
 
<math>\textbf{(A) }1440\qquad\textbf{(B) }2880\qquad\textbf{(C) }5760\qquad\textbf{(D) }182,440\qquad \textbf{(E) }362,880</math>
  
 +
=Solution=
 +
To solve this, treat the two Arabic books as one unit and the four Spanish books as another unit. Along with the three German books, you now have five units to arrange.
  
−
First lets see how we can arrange the first two books, then the next 4, then the next 5. What we are left with is (2!)(4!)(5!) gets us to 5760 ways.
+
You can arrange these five units in 5! ways. Within the Arabic block, the two books can be arranged in 2! ways, and within the Spanish block, the four books can be arranged in 4! ways.
  
−
==Solution 1==
+
Multiplying all these possibilities gives the total number of arrangements as:
−
Since the Arabic books and Spanish books have to be kept together, we can treat them both as just one book. That means we're trying to find the number of ways you can arrange one Arabic book, one Spanish book, and three German books, which is just <math>5</math> factorial. Now, we multiply this product by <math>2!</math> because there are <math>2!</math> ways to arrange the Arabic books within themselves, and <math>4!</math> ways to arrange the Spanish books within themselves. Multiplying all these together, we have <math>2! \cdot 4! \cdot 5!=\boxed{\textbf{(C) }5760}</math>.
+
 
 +
.
 +
 
 +
This results in 5,760 ways to arrange the nine books while keeping the Arabic and Spanish books together.
  
 
==Video Solution (CREATIVE ANALYSIS!!!)==
 
==Video Solution (CREATIVE ANALYSIS!!!)==

Latest revision as of 06:26, 17 October 2024

Problem

Professor Chang has nine different language books lined up on a bookshelf: two Arabic, three German, and four Spanish. How many ways are there to arrange the nine books on the shelf keeping the Arabic books together and keeping the Spanish books together?

$\textbf{(A) }1440\qquad\textbf{(B) }2880\qquad\textbf{(C) }5760\qquad\textbf{(D) }182,440\qquad \textbf{(E) }362,880$

Solution

To solve this, treat the two Arabic books as one unit and the four Spanish books as another unit. Along with the three German books, you now have five units to arrange.

You can arrange these five units in 5! ways. Within the Arabic block, the two books can be arranged in 2! ways, and within the Spanish block, the four books can be arranged in 4! ways.

Multiplying all these possibilities gives the total number of arrangements as:

.

This results in 5,760 ways to arrange the nine books while keeping the Arabic and Spanish books together.

Video Solution (CREATIVE ANALYSIS!!!)

https://youtu.be/GznYT3zMh3o

~Education, the Study of Everything

Video Solutions

https://youtu.be/ci5Nwd1UnNE

https://youtu.be/FqyYAiyjCds

~savannahsolver

https://www.youtube.com/watch?v=t1IARvN-JMM ~David

See Also

2018 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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 AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems/Problem_16&oldid=230078"