Problem 11 · 1991 AJHSME
Medium
Counting & Probability
count-pairs
There are several sets of three different numbers whose sum is 15 which can be chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. How many of these sets contain a 5?
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Answer: B — 4.
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Hint 1 of 3
The 5 is forced into the set, so it's already "spent." That fixes how much the other TWO numbers must add up to. Once you know that target, the question shrinks to counting pairs.
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Hint 2 of 3
Lock in the 5: the remaining two must total 15 − 5 = 10, be different from each other, and not be another 5. Count those pairs.
Still stuck? Show hint 3 →
Hint 3 of 3
List pairs from the small end (1 + 9, 2 + 8, …) and stop when you'd start repeating — once the two halves cross, you're just re-listing pairs you already have.
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Approach: fix the forced element, then count pairs for what's left
- Since 5 must be in the set, treat it as already chosen. The other two numbers then have to make up 15 − 5 = 10, using different values from 1–9 (and not a second 5).
- Walk up from the bottom: 1 + 9, 2 + 8, 3 + 7, 4 + 6. The next would be 5 + 5 (repeats 5, not allowed) and after that pairs just repeat in reverse. So there are 4 sets.
- Why this transfers: when one item is required, "use it up" first and reduce to a smaller problem on what remains — here a three-number sum collapses into an easy two-number pairing.
- Avoid double-counting: {5, 1, 9} and {5, 9, 1} are the same set, so list each pair only once by always taking the smaller number first.
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