🇺🇸 AMC 8 ⇄ switch contest
1996 AMC 8 Stretch

Problem 3

Problem 3 · AMC 8 Stretch Core
Counting & Probability and-process-multiplyconsidering-extreme-cases
Ms. Smith wants to tip her doorman. Her purse holds exactly four different things: a quarter, a half dollar, a silver dollar, and a five-dollar bill. She will give him some of these as a tip. How many different tips are possible (she does give at least one item)?
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Answer: 15 tips
Show hints
Hint 1 of 4
Treat each piece of money like a pizza topping: decide give it or keep it.
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Hint 2 of 4
There are 4 different items, each a yes/no choice. Multiplying the yes/no choices gives all the possible groups.
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Hint 3 of 4
\(2 \times 2 \times 2 \times 2\) counts every group, including the 'give nothing' group. But she definitely tips, so that one group is not allowed.
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Approach: AND process, then subtract the empty case
  1. Each of the 4 different items gets a give-it-or-not decision. That is a 4-step AND process: \(2 \times 2 \times 2 \times 2 = 2^4 = 16\).
  2. Those 16 groups include the 'give nothing' group. Since she definitely gives a tip, throw that one out: \(16 - 1 = 15\).
  3. Because all four items are different values, no two different groups are worth the same, so we are not over-counting. There are 15 possible tips.
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