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

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==Solution 5==  
 
==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 will be <math>200+45 = \boxed{\textbf{(B)}\ 245}</math> handshakes.  
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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)
 
-Benedict T (countmath1)
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 +
==Video Solution (HOW TO THINK CREATIVELY!!!)==
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https://youtu.be/TCxjwO-I8kQ
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 +
~Education, the Study of Everything
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==Video Solution==
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https://www.youtube.com/watch?v=PRF4Co5txJI
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 +
~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 (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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