Problem 13 · AMC 8 Stretch
Stretch
Counting & Probability
and-process-multiplyconsidering-extreme-caseslogical-reasoning
There are 3 different math books and 2 different art books, all different. Roslyn will choose some of them. How many selections can she make if (a) she takes at least one book; (b) she must include at least one math book?
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Answer: (a) 31; (b) 28
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Hint 1 of 4
All 5 books are different, so a selection is just a subset: each book is in or out. Start from \(2^5\).
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Hint 2 of 4
For 'at least one book', remove only the empty selection.
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Hint 3 of 4
For 'at least one math book', it is easiest to handle the math books and the art books separately. Math books: how many ways to pick 'at least one' of the 3? Art books: free to pick any.
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Approach: Subsets; subtract the empty case, and split into groups for part (b)
- All 5 books are different, so a selection is a subset; each book is in or out, giving \(2^5 = 32\) subsets in all.
- (a) At least one book: remove the empty selection, so \(2^5 - 1 = 31\).
- (b) At least one math book: split into the 3 math books and the 2 art books. Math must include at least one, so \(2^3 - 1 = 7\). Art is free, so \(2^2 = 4\).
- Multiply (AND process): \(7 \times 4 = 28\).
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