Difference between revisions of "2018 AMC 8 Problems/Problem 16"
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==Solution 1== | ==Solution 1== | ||
| − | 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>. | + | 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>. |
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==Video Solutions== | ==Video Solutions== | ||
Revision as of 09:34, 1 January 2023
Contents
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?
Solution 1
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 
factorial. Now, we multiply this product by 
because there are ways to arrange the Arabic books within themselves, and 
ways to arrange the Spanish books within themselves. Multiplying all these together, we have 
.
Video Solutions
~savannahsolver
https://www.youtube.com/watch?v=t1IARvN-JMM
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. 
