Difference between revisions of "2017 AMC 10A Problems/Problem 8"

m (Solution 3: LaTeX)
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We can focus on how many handshakes the <math>10</math> people who don't know anybody get.  
 
We can focus on how many handshakes the <math>10</math> people who don't know anybody get.  
  
The first person gets <math>29</math> handshakes.
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The first person gets <math>29</math> handshakes, the second person gets <math>28</math> handshakes, ..., and the tenth receives <math>20</math> handshakes.
 
 
Second gets <math>28</math>
 
 
 
......
 
 
 
And the tenth receives <math>20</math> handshakes.
 
  
 
We can write this as the sum of an arithmetic sequence.
 
We can write this as the sum of an arithmetic sequence.

Revision as of 23:45, 3 January 2020

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 hug, and people who do not know each other shake hands. How many handshakes occur within the group?

$\textbf{(A)}\ 240\qquad\textbf{(B)}\ 245\qquad\textbf{(C)}\ 290\qquad\textbf{(D)}\ 480\qquad\textbf{(E)}\ 490$

Solution 1

Each one of the ten people has to shake hands with all the $20$ other people they don’t know. So $10\cdot20 = 200$. 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 $\binom{10}{2} = 45$. Thus the answer is $200 + 45 = \boxed{\textbf{(B)}\ 245}$.

Solution 2

We can also use complementary counting. First of all, $\dbinom{30}{2}=435$ handshakes or hugs occur. Then, if we can find the number of hugs, then we can subtract it from $435$ to find the handshakes. Hugs only happen between the $20$ people who know each other, so there are $\dbinom{20}{2}=190$ hugs. $435-190= \boxed{\textbf{(B)}\ 245}$.

Solution 3

We can focus on how many handshakes the $10$ people who don't know anybody get.

The first person gets $29$ handshakes, the second person gets $28$ handshakes, ..., and the tenth receives $20$ handshakes.

We can write this as the sum of an arithmetic sequence.

$\frac{10(20+29)}{2}\implies 5(49)\implies 245.$ Therefore, the answer is $\boxed{\textbf{(B)}\ 245}$

See Also

2017 AMC 10A (ProblemsAnswer KeyResources)
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

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