What we see: a counting question. 6 couples make 12 people, and we choose 5 of them with no couple both on the committee.
Step 1. Turn the rule into a simple picture.
What we see: no couple can be on the committee together.
What it means: each couple can send at most one person. We need 5 people, so 5 different couples each send exactly one person, and 1 couple sends nobody.
What we do and why: count in two parts: which couples send someone, then which person from each.
Step 2. Choose the 5 couples.
What we see: 6 couples, and we pick 5 of them.
What it means: picking the 5 that send someone is the same as picking the 1 couple that sends nobody.
What we do and why: \(\binom{6}{5} = 6\) ways.
Step 3. Choose one person from each chosen couple.
What we see: each of the 5 chosen couples has 2 people.
What it means: each couple has 2 choices, and the choices do not depend on each other.
What we do and why: multiply: \(2 \times 2 \times 2 \times 2 \times 2 = 2^5 = 32\) ways.
Step 4. Put the parts together.
What we see: 6 ways for the couples, 32 ways for the people.
What we do and why: every couple choice works with every person choice, so multiply: \(6 \times 32 = 192\).
The trap: E (792) is \(\binom{12}{5}\), every committee of 5 with the couple rule ignored. A (32) forgets to choose which 5 couples are used.
Faster way: pick people one couple at a time: (couples to use) times (2 for each couple used).
Takeaway: When the rule is "no two from the same pair," first choose which pairs are used, then multiply by 2 for each pair.
Answer: B
