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Counting PUZZLE

Posted By: Bob Koca
Date: Friday, 3 May 2024, at 8:21 a.m.

In Response To: Counting PUZZLE (Nack Ballard)

In original position if you incorrectly counted 8/4 13/8 moving two different checkers as a different play than 13/4 moving a single checker you would be up to 17 moves.

In second position that duplication between plays involving a single checker being moved and plays moving two different checkers does not occur.

Here is a related question I include in my discrete math course. Suppose that a school has 6 Freshman, 7 Sophomores, 8 Juniors, and 9 Seniors. How many ways can they choose a committee of 5 students if it must have at least one from each grade?

Student A argues: We can first choose one of each grade in 6 x 7 x 8 x 9 = 3024 ways. We are now assured of having one of each grade. Having done that choose one of the remaining 26 students. This gives 3024 x 26 = 78624 ways.

Student B argues: We can break this into cases based on which grade has exactly 2 on the committee. This gives (6C2 x 7 x 8 x 9) + (6 x 7C2 x 8 x 9) + (6 x 7 x 8C2 x 9) + (6 x 7 x 8 x 9C2) = 7560 + 9072 + 10584 + 12096 = 39312 ways.

I have seen both arguments used by strong students. But at least one is clearly incorrect. What is wrong?

This semester a colleague and I have a sabbatical project of writing a discrete math text that will be used at our school. If anyone has familiarity with that area and wants to volunteer (I have nothing tangible to offer beyond a thanks and mention in the book) to look over a chapter and give comments let me know.

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