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The question is as follows: "A class of 10 students is electing a president and two vice presidents for student council. If each student in the class is eligible for either office and no student may hold multiple offices, how many different ways can the president and two vice presidents be selected?"

My approach this this problem was the following: 10 people can be president 9 people can be VP 1 8 people can be VP 2

Therefore, 1098 = 720 combinations.

The correct approach is: 10C1 people can be president 9C2 people can be VP

Therefore, 360 combinations exist.

My approach makes a lot more sense logically for me - can you help me understand what is incorrect about my approach and why the correct approach treats the 2 VP slots as 9C2 instead of: VP 1 = 9C1 VP 2 = 8C1

askakdagr8
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1 Answers1

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Order does not matter in a combination, the two vice presidents are indistinguishable. Your approach counts the vice presidencies as distinct positions, treating it as though it matters if someone is VP #1 or VP #2. Your approach double counts the number of combinations, since it treats having Persons A and B as vice presidents as different from having B and A. There are 720 ways to select and order three individuals from this group, but in this scenario, the order of the last two doesn't matter.

Nuclear Hoagie
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