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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>.
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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>.
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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==  
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==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 #4==  
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==Solution 4==  
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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 <math>20 * 10 = 200</math> handshakes. In addition, those 10 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.
+
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.
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 +
==Solution 5==
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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
 +
 
 +
==Video Solution==
 +
https://www.youtube.com/watch?v=PRF4Co5txJI
 +
 
 +
~Math4All999
  
 
== See Also ==
 
== See Also ==

Latest revision as of 06:22, 14 September 2024

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

Video Solution

https://www.youtube.com/watch?v=PRF4Co5txJI

~Math4All999

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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