Areteem: Math Challenge II-A Combinatorics

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Math Challenge II-A Combinatorics

Lecture 2, problem 2.24

You aren't given the names of the boys. The answer should be $6! \cdot 7 \cdot 6 \cdot 5 \cdot 4$.

Even if we aren't given the actual names in the problems, everyone does have a name. So when the problem says "the boys finish in alphabetical order based on their names" this means (as far as solving the problem is concerned) that the order of the boys is fixed and already known.

It looks like your solution wants to arrange the boys first, and then arrange the girls. Since the order of the boys is fixed, the $6!$ is not needed. You are then correct that there are $7$ spaces for the first girl. However, the problem doesn't say that girls can't finish next to each other, so there are actually $8$ spaces for the second girl. This pattern continues giving an answer of $1\cdot 7\cdot 8\cdot 9\cdot 10$ which matches the answer given.

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Suppose the first boy is A, the second is B, the third is C, the fourth is D, the fifth is E, and the sixth is F. There are $6!$ orderings of the boys, because boy A could be called Alex, and boy B could be called James, or boy A's name is Bill while boy B's name is Alan. There are two orderings, because boy A's mom could call him either Alex or Bill or any other name.

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I got it. Thanks. No more questions.

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