# 1996 AJHSME Problems/Problem 21

## Problem

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?

$\text{(A)}\ 6 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 10 \qquad \text{(D)}\ 12 \qquad \text{(E)}\ 24$

## Solution

To have an even sum with three numbers, we must add either $E+O+O$, or $E + E + E$, where $O$ represents an odd number, and $E$ represents an even number.

Since there are not three even numbers in the given set, $E+E+E$ is impossible. Thus, we must choose two odd numbers, and one even number.

There are $2$ choices for the even number.

There are $4$ choices for the first odd number. There are $3$ choices for the last odd number. But the order in which we pick these 2 numbers doesn't matter, so this overcounts the pairs of odd numbers by a factor of $2$. Thus, we have $\frac{4\cdot 3}{2} = 6$ choices for a pair of odd numbers.

In total, there are $2$ choices for an even number, and $6$ choices for the odd numbers, giving a total of $2\cdot 6 = 12$ possible choices for a 3-element set that has an even sum. This is option $\boxed{D}$.