Difference between revisions of "2018 AMC 8 Problems/Problem 16"

Revision as of 06:21, 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

4^8000000000 + 8^919888771 is the answer

Video Solution (CREATIVE ANALYSIS!!!)

https://youtu.be/GznYT3zMh3o

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Video Solutions

https://youtu.be/ci5Nwd1UnNE

https://youtu.be/FqyYAiyjCds

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

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.

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 ==Solution==
-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>.
+^8000000000 + 8^919888771 is the answer
 ==Video Solution (CREATIVE ANALYSIS!!!)==

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