1996 AMC 8 Problem 21

Below is the professionally curated solution for Problem 21 of the 1996 AMC 8, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1996 AMC 8 solutions, or check the answer key.

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Concepts:parity combinations

Difficulty rating: 1200

21.

How many subsets containing three different numbers can be selected from the set $\{89, 95, 99, 132, 166, 173\}$ so that the sum of the three numbers is even?

$6$

$8$

$10$

$12$

$24$

Solution:

The set has $4$ odd numbers ( $89, 95, 99, 173$ ) and $2$ even numbers ( $132, 166$ ). A sum of three is even only with two odds and one even, since three evens is impossible with just two available.

The count is $\binom{4}{2} \cdot \binom{2}{1} = 6 \cdot 2 = 12$ .

Thus, the correct answer is D .

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