Problem 6 · AMC 8 Stretch
Stretch
Counting & Probability
and-process-multiplycomplementary-counting
A club has 4 married couples (8 people total). A 3-person committee is chosen, but a husband and wife may NOT both be on it. How many committees are possible?
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Answer: 32 committees
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Hint 1 of 4
Since no two members can be a married pair, the 3 members must come from 3 DIFFERENT couples.
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Hint 2 of 4
Step 1: choose which 3 of the 4 couples will be represented. Step 2: from each chosen couple, pick the one person who serves (2 ways each).
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Hint 3 of 4
Choosing 3 couples out of 4 can be done in 4 ways (you leave out 1 couple). Then \(2 \times 2 \times 2\) for the one person from each. Multiply.
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Approach: Build in steps — choose the couples, then a person from each
- No two members can be spouses, so the 3 members come from 3 different couples. Build the committee in steps.
- Step 1: pick the 3 couples to be represented. With 4 couples, choosing 3 is the same as choosing the 1 couple to leave out, so there are 4 ways.
- Step 2: from each of the 3 chosen couples, pick which spouse serves — 2 ways each.
- Multiply: \(4 \times (2 \times 2 \times 2) = 4 \times 8 = 32\). There are 32 possible committees.
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