Problem 23 · 1991 AJHSME
Stretch
Counting & Probability
inclusion-exclusionvenn
The Pythagoras High School band has 100 female and 80 male members. The orchestra has 80 female and 100 male members. There are 60 females who are in both band and orchestra. Altogether there are 230 students who are in either band or orchestra or both. The number of males in the band who are NOT in the orchestra is
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Answer: A — 10.
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Hint 1 of 3
This looks tangled because males and females are mixed together. Untangle it: the 230 total splits cleanly into distinct females + distinct males. The females are fully solvable on their own (you know the band females, orchestra females, AND the overlap), so peel them off first.
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Hint 2 of 3
Find distinct females with overlap subtracted (band F + orchestra F − both F). Subtract from 230 to get distinct males. Then run the SAME overlap formula on the males to find how many are in both — and finally remove those from the band males.
Still stuck? Show hint 3 →
Hint 3 of 3
Once you know males-in-both, the answer is just band males minus that overlap (the ones left are in band only).
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Approach: handle females first, then inclusion-exclusion on the males
- Separate by gender. Distinct females (counting the 60 overlap only once): 100 + 80 − 60 = 120.
- The 230 total is females + males, so distinct males = 230 − 120 = 110.
- Now inclusion-exclusion on the males to find the overlap: (band males) + (orchestra males) − (distinct males) = 80 + 100 − 110 = 70 males are in both.
- Males in band but NOT orchestra = all band males minus those also in orchestra = 80 − 70 = 10.
- Why this transfers: when a population mixes two categories, split it into independent sub-populations you can fully solve, then apply "in A or B = A + B − both" to each. The overlap formula run backwards (solving for "both" from the union) is the key move here.
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