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

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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?
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==Problem==
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The Blue Bird High 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?
  
 
<math>\textbf{(A) }2\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad \textbf{(E) }12</math>
 
<math>\textbf{(A) }2\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad \textbf{(E) }12</math>
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==Solution==
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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>.
  
 
==Video Solution==
 
==Video Solution==
 
https://youtu.be/Zhsb5lv6jCI
 
https://youtu.be/Zhsb5lv6jCI
 
===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 = \boxed{\textbf{(E) }12}</math>.
 
  
 
==See Also==
 
==See Also==

Latest revision as of 23:53, 13 February 2021

Problem

The Blue Bird High 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 $2 \cdot 6 = \boxed{\textbf{(E) }12}$.

Video Solution

https://youtu.be/Zhsb5lv6jCI

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

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