2017 AMC 12A Problems/Problem 5

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.

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

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

Page

Toolbox

Search