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

Revision as of 18:35, 27 October 2020

Problem 16

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?

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

Video Solution

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.

@@ 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>
-==Solution==
+Video 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> factorial ways to arrange just the Arabic books, and <math>4!</math> ways to arrange just the Spanish books. Multiplying all these together, we have <math>2!\cdot 4!\cdot 5!=\boxed{\textbf{(C) }5760.}</math>
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Difference between revisions of "2018 AMC 8 Problems/Problem 16"

Revision as of 18:35, 27 October 2020

Problem 16

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