How it works
Complementary counting means
(what you want) = (total) − (what you don't want)
The phrase "at least one" is the big signal. "At least one even digit" splits into many cases, but its opposite, "no even digits at all", is a single easy count. The same works for probability: P(at least one) = 1 − P(none).
When a problem asks for at least one of two things, subtract each "missing" case and add back the overlap you removed twice (inclusion-exclusion): total − (no A) − (no B) + (neither).
Complementary Counting practice problems with solutions
These problems come from Mathyessy's question bank. Try each one before opening the solution.
How many 3 digit numbers from 100 to 999 have at least one even digit?
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Answer: 775
Count the opposite: numbers with all odd digits. Each of the three digits has 5 odd choices, giving 5 × 5 × 5 = 125. There are 900 three-digit numbers, so 900 − 125 = 775.
Flip 3 coins. Probability of at least one tail? Enter numerator over 8.
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Answer: 7/8
The only way to get no tails is HHH, with probability 1/8. So P(at least one tail) = 1 − 1/8 = 7/8.
Roll a die 5 times. Probability of at least one 6?
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Answer: 4651/7776
P(no 6 in 5 rolls) = (5/6)⁵ = 3125/7776. So P(at least one 6) = (7776 − 3125)/7776 = 4651/7776.
How many 3 digit numbers from 100 to 999 have at least one 5 and at least one 6?
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Answer: 52
No 5: 8 × 9 × 9 = 648 (the first digit cannot be 0 or 5). No 6: also 648. Neither 5 nor 6: 7 × 8 × 8 = 448. Inclusion-exclusion gives 900 − 648 − 648 + 448 = 52.
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Related topics
Related practice sets in the app
All Counting & Probability topics · Contest math practice by topic · MATHCOUNTS practice · AMC 8 practice