Problem 23 · AMC 8 Stretch
Stretch
Counting & Probability
and-process-multiplyor-process-add
How many different selections of at least one book can Erica make from 4 different books? (a) Solve it with an AND process. (b) Solve it with an OR process. Which way was easier?
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Answer: 15 selections (AND is easier)
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Hint 1 of 4
AND view: each of the 4 different books is independently in or out.
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Hint 2 of 4
OR view: split by how many books she takes — exactly 1, exactly 2, exactly 3, or exactly 4 — and add the counts.
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Hint 3 of 4
AND gives \(2^4\) minus the empty set. OR gives (ways to pick 1) + (pick 2) + (pick 3) + (pick 4). Both should match.
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Approach: Two ways to count the same total
- (a) AND process: each of the 4 different books is in or out — a 4-step AND process giving \(2^4 = 16\) subsets. 'At least one' removes the empty set: \(2^4 - 1 = 15\).
- (b) OR process: split by how many books she takes and add: \(4 + 6 + 4 + 1 = 15\) (pick 1, pick 2, pick 3, pick 4).
- Both give 15. The AND process is much easier — one quick power of 2 minus 1, instead of adding four separate counts.
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