Difference between revisions of "2017 AMC 10A Problems/Problem 8"
(→Solution 1) |
m (→Solution 2) |
||
| Line 9: | Line 9: | ||
==Solution 2== | ==Solution 2== | ||
| − | We can also use complementary counting. First of all, <math>\dbinom{30}{2}=435</math> handshakes or hugs occur. Then, if we can find the number of hugs, then we can subtract it from <math>435</math> to find the handshakes. Hugs only happen between the 20 people who know each other, so there are <math>\dbinom{20}{2}=190</math> hugs. <math>435-190= \boxed{\textbf{(B)}\ 245}</math>. | + | We can also use complementary counting. First of all, <math>\dbinom{30}{2}=435</math> handshakes or hugs occur. Then, if we can find the number of hugs, then we can subtract it from <math>435</math> to find the handshakes. Hugs only happen between the <math>20</math> people who know each other, so there are <math>\dbinom{20}{2}=190</math> hugs. <math>435-190= \boxed{\textbf{(B)}\ 245}</math>. |
==Solution 3== | ==Solution 3== | ||
Revision as of 19:18, 23 December 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 within the group?
Solution 1
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 
.
Solution 2
We can also use complementary counting. First of all, 
handshakes or hugs occur. Then, if we can find the number of hugs, then we can subtract it from 
to find the handshakes. Hugs only happen between the people who know each other, so there are 
hugs. 
.
Solution 3
We can focus on how many handshakes the 10 people get.
The 1st person gets 29 handshakes.
2nd gets 28
......
And the 10th receives 20 handshakes.
We can write this as the sum of an arithmetic sequence.
Therefore, 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. 
