Thanks for the comment Jan! I’ve added several sentences leading up to why that is true:

“Well, given a particular permutation, say ABACAA, there are a number of ways we could move the As around… The same number in fact as the number of ways we could permute A₁A₂A₃A₄: 4!, the factorial of 4. Now, if we had distinguished between the As, there would have been 6! ways to arrange ABACAA. So, including the clones, we have 6! possibilities, with 4! copies of every unique one. Dividing by the number of duplicates then allows us to get rid of them. So in this case, we divide by 4!, this gives us 6!/4!=6*5=30.”

>Just mention the, that the pyramid is explained in exercise B…

Done 🙂

]]>Thanks for the feedback!

I’ve rewritten this as: “Note that the value of the element in the th position of the th row is in the above diagram. This is true in general.”

I didn’t really need them to realize why it was true at that point, just notice that it was. Then I could explain it 🙂

]]>In the same context I would also say:

“To cancel out the clones, we divide by 4!,” is not so obvious to readers that are new to the problem.

In exercise D you refer to “the pyramide”. If one (like me) has jumped over this exercise he would be confused about this fact. Just mention the, that the pyramid is explained in exercise B…

Jan

]]>Your questions are great explorations. Thanks for sharing.

The only constructive criticism I’d offer from my time as a teacher regards your sentence between the first two images: “One quickly notices that the value of the element in the position of the row is .” As mathematicians, I agree that we see that connection, but almost all of the students I’ve taught have had to be coached into that realization. It is a critical connection, but I don’t think it’s a quick notice for very many

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