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2017 AMC 10A Problem 8

2017 AMC 10A Problem 8AMC 10 Step by Step

Name: Date: September 13, 2026

Basic Counting · difficulty 2 of 5 · about 2 min · position P1-10

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?

A)
$240$
B)
$245$
C)
$290$
D)
$480$
E)
$490$

Answer

Key insight

A handshake happens exactly when at least one of the pair is a stranger: 10 times 20 mixed pairs plus C(10,2) stranger-stranger pairs.

Solution

The only pairs who know each other are pairs drawn from the group of $20$ . Every other pair shakes hands, so a handshake occurs exactly when the pair contains at least one of the $10$ strangers.

Pairs with one stranger and one member of the $20$ : $10 \cdot 20 = 200$ .

Pairs of two strangers: $\binom{10}{2} = 45$ .

Total handshakes: $200 + 45 = 245$ .

The answer is $\boxed{\textbf{(B)}\ 245}$ .

Why this works

Split the pairs by how many strangers they contain. Each pair is either a hug (both in the acquainted group) or a handshake (everything else), so the classification is exhaustive and non-overlapping. Counting "one from each group" is a product; counting "two from the same group" is a combination.

Alternative approach

Complementary counting: there are $\binom{30}{2} = 435$ pairs altogether, and $\binom{20}{2} = 190$ of them hug. The rest shake hands: $435 - 190 = 245$ .

The trap

Forgetting that the 10 strangers also shake hands with each other, giving only 200, or counting each handshake twice, giving 490.

Common mistakes

Forgetting that the 10 strangers also shake hands with each other, giving only 200, or counting each handshake twice, giving 490.
Using $10 \cdot 29 = 290$ , which counts each stranger-stranger handshake twice.

Techniques

Count the complement and subtract from the total · Set up the equation/formula and compute; no special trick needed