Problem 21 · 1996 AJHSME
Hard
Counting & Probability
paritycombinations
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?
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Answer: D — 12.
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Hint 1 of 2
The actual values don't matter — replace every number by just 'odd' or 'even.' The set is really four odds and two evens. Whether a sum is even depends only on how many odd numbers you grabbed.
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Hint 2 of 2
Adding odds is what flips even/odd: a sum is even only when you use an EVEN number of odd terms. Picking 3 numbers, that means 2 odds + 1 even (using 0 odds would need 3 evens, but there are only 2). Now just count the ways.
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Approach: reduce to parity, then count
- Tag each number: odds are 89, 95, 99, 173 (four of them), evens are 132, 166 (two). A total is even only when an even count of the picks are odd. With 3 picks the only workable split is 2 odds + 1 even — the alternative '0 odds' would need 3 evens, but only 2 exist.
- Count those choices: pick 2 of the 4 odds in C(4,2) = 6 ways, and 1 of the 2 evens in C(2,1) = 2 ways. Total 6 · 2 = 12 subsets.
- Why this transfers: for even/odd-sum counting, throw away the numbers and keep only odd/even labels. Even sum = even number of odd parts — a rule that turns a scary value problem into simple counting.
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