Difference between revisions of "2017 AMC 12A Problems/Problem 5"
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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> | ||
− | =Alternate solution== | + | ==Alternate solution== |
− | + | 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> | |
== See Also == | == See Also == |
Revision as of 15:28, 9 February 2017
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
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
Alternate solution
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.