Difference between revisions of "2017 AMC 12A Problems/Problem 5"
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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 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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| + | <math>29+28+27+26+25+24+23+22+21+20 = \boxed{(B)=\ 245}</math> | ||
==Solution== | ==Solution== | ||
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==Solution - Complementary Counting== | ==Solution - 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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== See Also == | == See Also == | ||
Revision as of 01:07, 6 January 2019
Problem
At a gathering of 
people, there are 
people who all know each other and 
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?
Solution - Basic
All of the handshakes will involve at least one person from the who know no one. Label these ten people 
, 
, 
, 
, 
, 
, 
, 
, 
, 
.
Person from the group of 10 will initiate a handshake with everyone else (
people). Person initiates 
handshakes plus the one already counted from person . Person initiates 
new handshakes plus the two we already counted. This continues until person initiates handshakes plus the nine we already counted from ... .
Solution
Let the group of people who all know each other be , and let the group of people who know no one be . Handshakes occur between each pair 
such that 
and 
, and between each pair of members in . Thus, the answer is
Solution - 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 
and 
, respectively. Thus, the total amount of handshakes is 
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
