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Difference between revisions of "2017 AMC 12A Problems/Problem 5"

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==Alternate solution==  
 
==Alternate solution==  
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The number of handshakes will be equivalent to the difference between the number of total interactions and the number of hugs, which are  <math>{30\choose 2}</math> and <math>{20\choose 2}</math>, respectively. Thus the total amount of handshakes  is <math>{30\choose 2} - {20\choose 2} = 435 - 190= \boxed{(B)=\ 245}  </math>
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The number of handshakes will be equivalent to the difference between the number of total interactions and the number of hugs, which are  <math>{30\choose 2}</math> and <math>{20\choose 2}</math>, respectively. Thus, the total amount of handshakes  is <math>{30\choose 2} - {20\choose 2} = 435 - 190= \boxed{(B)=\ 245}  </math>
  
 
== See Also ==
 
== See Also ==

Revision as of 15:35, 18 February 2017

Problem

At a gathering of $30$ people, there are $20$ people who all know each other and $10$ people who know no one. People who know each other hug, and people who do not know each other shake hands. How many handshakes occur?

$\textbf{(A)}\ 240\qquad\textbf{(B)}\ 245\qquad\textbf{(C)}\ 290\qquad\textbf{(D)}\ 480\qquad\textbf{(E)}\ 490$

Solution

Let the group of people who all know each other be $A$, and let the group of people who know no one be $B$. Handshakes occur between each pair $(a,b)$ such that $a\in A$ and $b\in B$, and between each pair of members in $B$. Thus, the answer is

$|A||B|+{|B|\choose 2} = 20\cdot 10+{10\choose 2} = 200+45 = \boxed{(B)=\ 245}$

Alternate solution

The number of handshakes will be equivalent to the difference between the number of total interactions and the number of hugs, which are ${30\choose 2}$ and ${20\choose 2}$, respectively. Thus, the total amount of handshakes is ${30\choose 2} - {20\choose 2} = 435 - 190= \boxed{(B)=\ 245}$

See Also

2017 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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 AMC 10 Problems and Solutions
2017 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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 AMC 12 Problems and Solutions

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