1990 AMC 12 Problem 16

Attempt Problem 16 of the 1990 AMC 12 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 1990 AMC 12 solutions, or check the answer key.

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

16.

At one of George Washington’s parties, each man shook hands with everyone except his spouse, and no handshakes took place between women. If 1313 married couples attended, how many handshakes were there among these 2626 people?
7878
185185
234234
312312
325325
Answer: C
Concepts:combinationscomplementary counting
Small hint:
Begin with all unordered pairs of the 2626 guests
Big hint:
Remove spouse pairs and pairs consisting of two women
Solution:
There are (262)=325\binom{26}{2}=325 possible pairs. Exclude the 1313 married pairs and the (132)=78\binom{13}{2}=78 pairs of women. The number of handshakes is 325−13−78=234. 325-13-78=234.
Thus the correct answer is C.
LIVE by Po-Shen Loh
Programs

LIVE Classes

Self-Paced Classes

Stars

Scholarship

About Us

Who Is Po-Shen Loh?

Email Contact

Newsletter

Jobs

Become a Star

For a more Thought Full world.

© 2026 Expii, Inc.