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Difference between revisions of "2017 AMC 12A Problems/Problem 5"
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<math>\textbf{(A)}\ 240\qquad\textbf{(B)}\ 245\qquad\textbf{(C)}\ 290\qquad\textbf{(D)}\ 480\qquad\textbf{(E)}\ 490</math>
 
<math>\textbf{(A)}\ 240\qquad\textbf{(B)}\ 245\qquad\textbf{(C)}\ 290\qquad\textbf{(D)}\ 480\qquad\textbf{(E)}\ 490</math>
  
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==Solution - Basic==
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==Solution 1 (Basic)==
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All of the handshakes will involve at least one person from the <math>10</math> who know no one.  Label these ten people <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math>, <math>E</math>, <math>F</math>, <math>G</math>, <math>H</math>, <math>I</math>, <math>J</math>.
+
All of the handshakes will involve at least one person from the <math>10</math> who knows no one.  Label these ten people <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math>, <math>E</math>, <math>F</math>, <math>G</math>, <math>H</math>, <math>I</math>, <math>J</math>.
  
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Person <math>A</math> from the group of 10 will initiate a handshake with everyone else (<math>29</math> people).  Person <math>B</math> initiates <math>28</math> handshakes plus the one already counted from person <math>A</math>.  Person <math>C</math> initiates <math>27</math> new handshakes plus the two we already counted.  This continues until person <math>J</math> initiates <math>20</math> handshakes plus the nine we already counted from <math>A</math> ... <math>I</math>.
+
Person <math>A</math> from the group of <math>10</math> will initiate a handshake with everyone else (<math>29</math> people).  Person <math>B</math> initiates <math>28</math> handshakes plus the one already counted from person <math>A</math>.  Person <math>C</math> initiates <math>27</math> new handshakes plus the two we already counted.  This continues until person <math>J</math> initiates <math>20</math> handshakes plus the nine we already counted from <math>A</math> ... <math>I</math>.
  
 
<math>29+28+27+26+25+24+23+22+21+20 = \boxed{(B)=\ 245}</math>
 
<math>29+28+27+26+25+24+23+22+21+20 = \boxed{(B)=\ 245}</math>
  
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==Solution==
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==Solution 2==
  
 
Let the group of people who all know each other be <math>A</math>, and let the group of people who know no one be <math>B</math>. Handshakes occur between each pair <math>(a,b)</math> such that <math>a\in A</math> and <math>b\in B</math>, and between each pair of members in <math>B</math>. Thus, the answer is
 
Let the group of people who all know each other be <math>A</math>, and let the group of people who know no one be <math>B</math>. Handshakes occur between each pair <math>(a,b)</math> such that <math>a\in A</math> and <math>b\in B</math>, and between each pair of members in <math>B</math>. Thus, the answer is
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<math>|A||B|+{|B|\choose 2} = 20\cdot 10+{10\choose 2} = 200+45 = \boxed{(B)=\ 245}</math>
 
<math>|A||B|+{|B|\choose 2} = 20\cdot 10+{10\choose 2} = 200+45 = \boxed{(B)=\ 245}</math>
  
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==Solution - Complementary Counting==  
+
==Solution 3 (Complementary Counting)==  
 
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>
 
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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==Solution==  
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==Solution 4==  
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Each person who does not know anybody will shake hands with all 20 of the people who know each other. This means there will be at least <math>20 * 10 = 200</math> handshakes. In addition, the entire group of people who don't know anyone will shake hands with each other, giving another <math>9+8+7+6+5+4+3+2+1 = 45</math> handshakes. Therefore, there is a total of $200+45 = 245 handshakes \boxed{(B)}\$.
+
Each of the <math>10</math> people who do not know anybody will shake hands with all <math>20</math> of the people who do know each other. This means there will be at least <math>20 * 10 = 200</math> handshakes. In addition, those <math>10</math> people will also shake hands with each other, giving us another <math>9+8+7+6+5+4+3+2+1 = 45</math> handshakes. Therefore, there is a total of <math>200+45 = \boxed{(B) = 245}</math> handshakes.
 +
 
 +
==Solution 5==
 +
Every one of the <math>20</math> people who know each other will shake hands with each of the <math>10</math> people who know no one, so there are <math>20\cdot 10 = 200</math> handshakes here. Each of the <math>10</math> people will also shake hands with each other, so there will be <math>{10 \choose 2}</math> <math>=45</math> handshakes for this case. In total, there are <math>200+45 = \boxed{\textbf{(B)}\ 245}</math> handshakes.
 +
 
 +
-Benedict T (countmath1)
 +
 
 +
==Video Solution (HOW TO THINK CREATIVELY!!!)==
 +
https://youtu.be/TCxjwO-I8kQ
 +
 
 +
~Education, the Study of Everything
 +
 
 
== See Also ==
 
== See Also ==
 
{{AMC10 box|year=2017|ab=A|num-b=7|num-a=9}}
 
{{AMC10 box|year=2017|ab=A|num-b=7|num-a=9}}
 
{{AMC12 box|year=2017|ab=A|num-b=4|num-a=6}}
 
{{AMC12 box|year=2017|ab=A|num-b=4|num-a=6}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 15:40, 9 June 2023

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 1 (Basic)

All of the handshakes will involve at least one person from the $10$ who knows no one. Label these ten people $A$, $B$, $C$, $D$, $E$, $F$, $G$, $H$, $I$, $J$.

Person $A$ from the group of $10$ will initiate a handshake with everyone else ($29$ people). Person $B$ initiates $28$ handshakes plus the one already counted from person $A$. Person $C$ initiates $27$ new handshakes plus the two we already counted. This continues until person $J$ initiates $20$ handshakes plus the nine we already counted from $A$ ... $I$.

$29+28+27+26+25+24+23+22+21+20 = \boxed{(B)=\ 245}$

Solution 2

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

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

Solution 3 (Complementary Counting)

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 ${30\choose 2} - {20\choose 2} = 435 - 190= \boxed{(B)=\ 245}$

Solution 4

Each of the $10$ people who do not know anybody will shake hands with all $20$ of the people who do know each other. This means there will be at least $20 * 10 = 200$ handshakes. In addition, those $10$ people will also shake hands with each other, giving us another $9+8+7+6+5+4+3+2+1 = 45$ handshakes. Therefore, there is a total of $200+45 = \boxed{(B) = 245}$ handshakes.

Solution 5

Every one of the $20$ people who know each other will shake hands with each of the $10$ people who know no one, so there are $20\cdot 10 = 200$ handshakes here. Each of the $10$ people will also shake hands with each other, so there will be ${10 \choose 2}$ $=45$ handshakes for this case. In total, there are $200+45 = \boxed{\textbf{(B)}\ 245}$ handshakes.

-Benedict T (countmath1)

Video Solution (HOW TO THINK CREATIVELY!!!)

https://youtu.be/TCxjwO-I8kQ

~Education, the Study of Everything

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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