Engineering School Winter Party: Solving Tricky Combinations Question

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The discussion revolves around calculating the number of possible attendee combinations for a Winter Solstice party hosted by the dean of an Engineering School, involving 9 department chairpersons and their spouses. Each chair can attend with or without their spouse, but spouses cannot attend without their chair. Participants initially analyze the scenario for one and two chairs to build understanding before generalizing to nine. The complexity arises from the varying attendance combinations of chairs and their spouses. Ultimately, the problem highlights the combinatorial nature of the situation, making it a challenging yet engaging mathematical exercise.
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The dean of an Engineering School invites all 9 department chairpersons and their spouses to his house for a Winter Solstice party. If each department chair may attend without a spouse, but the spouse may not attend without the department chair, how many different sets of attendees are possible?

Anyone know the logic to set this answer up?
 
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Try it for the case of 1 department chair. Now do 2. You can count those pretty easily. Now do you see how to get to 9?
 
Forget about the fact that there are 9 of them, for the moment, and just focus on one chair. How many ways can the chair be represented at the event?

EDIT: I see Dick beat me to it
 
Actually, for a precaclulus math class, that can be a pretty complex problem, since for each possible set of department chairs that can show up, their corresponding spouses may or may not necessarily show up as well.
 
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