2017 AMC 10A Problems/Problem 8

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.

Second gets $28$

......

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 $\boxed{\textbf{(B)}\ 245}$

2017 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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2017 AMC 10A Problems/Problem 8

Contents

Problem

Solution 1

Solution 2

Solution 3

See Also