Difference between revisions of "2017 AMC 10A Problems/Problem 8"
m |
m (→Solution) |
||
| Line 6: | Line 6: | ||
==Solution== | ==Solution== | ||
| − | 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</math> | + | 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, or <math>\binom{10}{2} = 45</math>. Thus the answer is <math>200 + 45 = \boxed{\textbf{(B)} 245}</math>. |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2017|ab=A|num-b=7|num-a=9}} | {{AMC10 box|year=2017|ab=A|num-b=7|num-a=9}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 17:27, 8 February 2017
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?
Solution
Each one of the ten people has to shake hands with all the 
other people they don’t know. So 
. 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, or 
. Thus the answer 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
