2017 AMC 10A Problem 8

Every pair of people either hugs or shakes hands. The only pairs who hug are two people from the group of $20$ friends. So a pair shakes hands exactly when at least one of them is a stranger. Count those pairs in two groups.

1. One stranger and one friend. Each of the $10$ strangers shakes hands with each of the $20$ friends:
$$10 \times 20 = 200 \text{ handshakes}.$$
2. Two strangers. The strangers do not know each other either, so they shake hands too. Each of the $10$ strangers shakes $9$ other strangers' hands, giving $10 \times 9 = 90$. But that counts every handshake twice (once from each person), so divide by $2$:
$$\frac{10 \times 9}{2} = 45 \text{ handshakes}.$$
3. Total: $200 + 45 = 245$.

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

Sort the pairs by how many strangers they contain: zero (a hug), one, or two. Every pair falls into exactly one of these, so nothing is missed or double counted. "One from each group" is a simple multiplication, while "two from the same group" needs the divide-by-2 step because each pair gets counted from both ends.

Count all pairs and take away the hugs. With $30$ people there are $\frac{30 \times 29}{2} = 435$ pairs. The hugging pairs are pairs of friends: $\frac{20 \times 19}{2} = 190$. Everyone else shakes hands: $435 - 190 = 245$.

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