1996 AMC 8 Problem 21

Question

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

With only two even numbers available, an even sum must use two odd numbers and one even

There are

4

odd and

2

even numbers; choose

2

of the

4

odds and

1

of the

2

evens

The set has

4

odd numbers—

89,

95,

99,

and

173

—and

2

even numbers—

132

and

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 .

Problem 21 in other years