Difference between revisions of "2015 AMC 8 Problems/Problem 4"

m (Solution)
Line 4: Line 4:
  
 
===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>}.
 +
 
 
==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 13:04, 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?

$\textbf{(A) }2\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad \textbf{(E) }12$

Solution

There are $2$ ways to order the boys on the end, and there are $3!=6$ 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 (ProblemsAnswer KeyResources)
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. AMC logo.png