Difference between revisions of "2018 AMC 8 Problems/Problem 16"
(→Solution) |
(→Solution) |
||
| Line 5: | Line 5: | ||
==Solution== | ==Solution== | ||
| − | Since the Arabic books and Spanish books have to by 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>. | + | Since the Arabic books and Spanish books have to by 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>.-xyab |
==See Also== | ==See Also== | ||
Revision as of 12:07, 24 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?
Solution
Since the Arabic books and Spanish books have to by 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 
factorial ways to arrange just the Arabic books, and 
ways to arrange just the Spanish books. Multiplying all these together, we have 
.-xyab
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. 
