Difference between revisions of "2017 AMC 12A Problems/Problem 5"
m (LaTeXed numbers and made titles consistent) |
Countmath1 (talk | contribs) (→Solution 4) |
||
Line 23: | Line 23: | ||
==Solution 4== | ==Solution 4== | ||
Each of the <math>10</math> people who do not know anybody will shake hands with all <math>20</math> of the people who do know each other. This means there will be at least <math>20 * 10 = 200</math> handshakes. In addition, those <math>10</math> people will also shake hands with each other, giving us another <math>9+8+7+6+5+4+3+2+1 = 45</math> handshakes. Therefore, there is a total of <math>200+45 = \boxed{(B) = 245}</math> handshakes. | Each of the <math>10</math> people who do not know anybody will shake hands with all <math>20</math> of the people who do know each other. This means there will be at least <math>20 * 10 = 200</math> handshakes. In addition, those <math>10</math> people will also shake hands with each other, giving us another <math>9+8+7+6+5+4+3+2+1 = 45</math> handshakes. Therefore, there is a total of <math>200+45 = \boxed{(B) = 245}</math> handshakes. | ||
+ | |||
+ | ==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. | ||
+ | |||
+ | -Benedict T (countmath1) | ||
== See Also == | == See Also == |
Revision as of 22:09, 25 April 2023
Contents
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 1 (Basic)
All of the handshakes will involve at least one person from the who knows no one. Label these ten people
,
,
,
,
,
,
,
,
,
.
Person from the group of
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 2
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 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 and
, respectively. Thus, the total amount of handshakes is
Solution 4
Each of the people who do not know anybody will shake hands with all
of the people who do know each other. This means there will be at least
handshakes. In addition, those
people will also shake hands with each other, giving us another
handshakes. Therefore, there is a total of
handshakes.
Solution 5
Every one of the people who know each other will shake hands with each of the
people who know no one, so there are
handshakes here. Each of the
people will also shake hands with each other, so there will be
handshakes for this case. In total, there will be
handshakes.
-Benedict T (countmath1)
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.