Difference between revisions of "2015 AMC 8 Problems/Problem 4"
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===Solution=== | ===Solution=== | ||
| − | There are <math>2</math> ways to order the boys on the end, and there are <math>3!=6</math> ways to order the girls in the middle. We get the answer to be <math>2 \cdot 6 = \textbf{(E) }12</math>. | + | There are <math>2</math> ways to order the boys on the end, and there are <math>3!=6</math> ways to order the girls in the middle. We get the answer to be <math>2 \cdot 6 = \boxed{\textbf{(E) }12</math>}. |
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==See Also== | ==See Also== | ||
{{AMC8 box|year=2015|num-b=3|num-a=5}} | {{AMC8 box|year=2015|num-b=3|num-a=5}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 12:04, 26 November 2015
The Centerville Middle School chess team consists of two boys and three girls. A photographer wants to take a picture of the team to appear in the local newspaper. She decides to have them sit in a row with a boy at each end and the three girls in the middle. How many such arrangements are possible?
Solution
There are 
ways to order the boys on the end, and there are 
ways to order the girls in the middle. We get the answer to be $2 \cdot 6 = \boxed{\textbf{(E) }12$ (Error compiling LaTeX. Unknown error_msg)}.
See Also
| 2015 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 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. 
