2017 AMC 10A Problem 8

Attempt Problem 8 of the 2017 AMC 10A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2017 AMC 10A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

8.

At a gathering of 3030 people, there are 2020 people who all know each other and 1010 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?
240240
245245
290290
480480
490490
Answer: B
Concepts:counting pairscombinations
Small hint:
Count handshakes involving the ten people who know no one
Big hint:
Avoid double-counting by thinking in pairs
Solution:
Each of the 1010 people shake hands with each of the 2020 people. This results in 10⋅20=20010 \cdot 20 = 200 handshakes.
There are also (102)=45\binom{10}{2} = 45 handshakes within the 1010 people (every pair of people shake hands).
Therefore, the total number of handshakes is 200+45=245.200 + 45 = 245.
Thus, B is the correct answer.
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