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

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<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>
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==Solution==
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There are <math>2!</math> ways to arrange the Arabic books within their block, <math>4!</math> for Spanish, and <math>5!</math> for the two blocks and three books, for a product of <math>2!4!5!=\boxed{5760} \textbf{(C)}</math>.
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{{AMC8 box|year=2018|num-b=15|num-a=17}}
 
{{AMC8 box|year=2018|num-b=15|num-a=17}}
  
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 17:50, 21 November 2018

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?

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

Solution

There are $2!$ ways to arrange the Arabic books within their block, $4!$ for Spanish, and $5!$ for the two blocks and three books, for a product of $2!4!5!=\boxed{5760} \textbf{(C)}$.

2018 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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All AJHSME/AMC 8 Problems and Solutions

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