Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2017 AMC 10A Problems/Problem 8"

(→‎Solution)
 
(28 intermediate revisions by 19 users not shown)
Line 1: Line 1:
 
==Problem==
 
==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 a hug, and people who do not know each other shake hands. How many handshakes occur?
+
At a gathering of <math>30</math> people, there are <math>20</math> people who all know each other and <math>10</math> 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 within the group?
  
 
<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>
  
−
==Solution==
+
==Solution 1==
−
Each one of the ten people has to shake hands with all the 20 other people they don’t know. So <math>10\times 20</math> = 200. From there you also have to calculate how many handshakes occurred between the people who don’t know each other. Each person out of the 10 has to shake hands with 9 other people. That’s 90, however you have to take into account the overlap. For example, there's 10 people so...
+
Each one of the ten people has to shake hands with all the <math>20</math> other people they don’t know. So <math>10\cdot20 = 200</math>. From there, we calculate how many handshakes occurred between the people who don’t know each other. This is simply counting how many ways to choose two people to shake hands from <math>10</math>, or <math>\binom{10}{2} = 45</math>. Thus the answer is <math>200 + 45 = \boxed{\textbf{(B)}\ 245}</math>.
  
−
Person 1 shakes hands with people- 2,3,4,5,6,7,8,9,10
+
==Solution 2==
−
And person 2 shakes hands with people- 1,3,4,5,6,7,8,9,10 and so on.  
+
We can also use complementary counting. First of all, <math>\dbinom{30}{2}=435</math> handshakes or hugs occur. Then, if we can find the number of hugs, then we can subtract it from <math>435</math> to find the handshakes. Hugs only happen between the <math>20</math> people who know each other, so there are <math>\dbinom{20}{2}=190</math> hugs. <math>435-190= \boxed{\textbf{(B)}\ 245}</math>.
  
−
There's overlap. Person 1 shakes hands with person 2 and then person 2 shakes hands with person 1. So 90 needs to be divided by 2 so the overlap is taken into account. From there, add 200 + 45 to get the answer. (B) 245.
+
==Solution 3==
 +
We can focus on how many handshakes the <math>10</math> people who don't know anybody get.
 +
 
 +
The first person gets <math>29</math> handshakes with other people not him/herself, the second person gets <math>28</math> handshakes with other people not him/herself and not the first person, ..., and the tenth receives <math>20</math> handshakes with other people not him/herself and not the first, second, ..., ninth person. We can write this as the sum of an arithmetic sequence:
 +
 
 +
<math>\frac{10(20+29)}{2}\implies 5(49)\implies 245.</math>
 +
Therefore, the answer is <math>\boxed{\textbf{(B)}\ 245}</math>
 +
 
 +
 
 +
==Solution 4==
 +
First, we can find out the number of handshakes that the <math>10</math> people who don't know anybody share with the <math>20</math> other people. This is simply <math>10 \cdot 20 = 200</math>. Next, we need to find out the number of handshakes that are shared within the <math>10</math> people who don't know anybody. Here, we can use the formula <math>\frac{n(n-1)}{2}</math>, where <math>n</math> is the number of people being counted. The reason we divide by <math>2</math> is because <math>n(n-1)</math> counts the case where the <math>1^{st}</math> person shakes hands with the <math>2^{nd}</math> person <math>and</math> the case where the <math>2^{nd}</math> shakes hands with the <math>1^{st}</math> (and these 2 cases are the same). Thus, plugging <math>n=10</math> gives us <math>\frac{10 \cdot 9}{2} \implies 45</math>. Adding up the 2 cases gives us <math>200+45=\boxed{\textbf{(B)}\ 245}</math>
 +
 
 +
 
 +
== Video Solution ==
 +
https://youtu.be/3MiGotKnC_U?t=1627
 +
~ ThePuzzlr
 +
 
 +
 
 +
==Video Solution==
 +
https://youtu.be/pxg7CroAt20
 +
 
 +
https://youtu.be/cTtqZmui7D4
 +
 
 +
~savannahsolver
 +
 
 +
== Video Solution ==
 +
https://youtu.be/0W3VmFp55cM?t=3464
 +
 
 +
~ pi_is_3.14
 +
 
 +
==See Also==
 +
{{AMC10 box|year=2017|ab=A|num-b=7|num-a=9}}
 +
{{MAA Notice}}
 +
 
 +
[[Category:Introductory Combinatorics Problems]]

Latest revision as of 11:47, 19 October 2021

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 within the group?

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

Solution 1

Each one of the ten people has to shake hands with all the $20$ other people they don’t know. So $10\cdot20 = 200$. From there, we calculate how many handshakes occurred between the people who don’t know each other. This is simply counting how many ways to choose two people to shake hands from $10$, or $\binom{10}{2} = 45$. Thus the answer is $200 + 45 = \boxed{\textbf{(B)}\ 245}$.

Solution 2

We can also use complementary counting. First of all, $\dbinom{30}{2}=435$ handshakes or hugs occur. Then, if we can find the number of hugs, then we can subtract it from $435$ to find the handshakes. Hugs only happen between the $20$ people who know each other, so there are $\dbinom{20}{2}=190$ hugs. $435-190= \boxed{\textbf{(B)}\ 245}$.

Solution 3

We can focus on how many handshakes the $10$ people who don't know anybody get.

The first person gets $29$ handshakes with other people not him/herself, the second person gets $28$ handshakes with other people not him/herself and not the first person, ..., and the tenth receives $20$ handshakes with other people not him/herself and not the first, second, ..., ninth person. We can write this as the sum of an arithmetic sequence:

$\frac{10(20+29)}{2}\implies 5(49)\implies 245.$ Therefore, the answer is $\boxed{\textbf{(B)}\ 245}$


Solution 4

First, we can find out the number of handshakes that the $10$ people who don't know anybody share with the $20$ other people. This is simply $10 \cdot 20 = 200$. Next, we need to find out the number of handshakes that are shared within the $10$ people who don't know anybody. Here, we can use the formula $\frac{n(n-1)}{2}$, where $n$ is the number of people being counted. The reason we divide by $2$ is because $n(n-1)$ counts the case where the $1^{st}$ person shakes hands with the $2^{nd}$ person $and$ the case where the $2^{nd}$ shakes hands with the $1^{st}$ (and these 2 cases are the same). Thus, plugging $n=10$ gives us $\frac{10 \cdot 9}{2} \implies 45$. Adding up the 2 cases gives us $200+45=\boxed{\textbf{(B)}\ 245}$


Video Solution

https://youtu.be/3MiGotKnC_U?t=1627 ~ ThePuzzlr


Video Solution

https://youtu.be/pxg7CroAt20

https://youtu.be/cTtqZmui7D4

~savannahsolver

Video Solution

https://youtu.be/0W3VmFp55cM?t=3464

~ pi_is_3.14

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_8&oldid=163755"