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I'm working my way through the Section 5.4 exercises for the Grimaldi textbook, and the book's answer to one of the exercises doesn't make any sense to me. I was hoping someone could help me understand how it reaches its given answer.

Here's the problem: Let $|A| = 5$. How many closed binary operations on $A$ are commutative?

The book gives the answer $5^{10}$, but that makes little sense to me. When I do the problem, I get $5^{15}$.

Here's how I reach $5^{15}$. There are $5$ functions that have double pairs (i.e., $f(a,a),f(b,b),f(c,c),f(d,d),f(e,e)$), and each of those has $5$ possible answers. Thus, there are $5^5$ potential arrangements for those $5$ inputs. For the rest of the inputs (which should be only $20$, since $25-5=20$), each combination of inputs corresponds with another combination if the function is commutative (i.e., $f(a,b) = f(b,a)$). So we only have to match half (i.e., $10$) of the remaining possible combinations with answers. That leaves us with $5^{10}$ potential arrangements for the remaining $10$ possible inputs. The total number of commutative operations should therefore be $5^{5} \times 5^{10} = 5^{15}$, not $5^{10}$.

Given this discrepancy, I thought I'd share my thoughts for scrutiny and see if I'm missing something. Any help or insight on this would be greatly appreciated. Thank you!

mike_peta
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    I believe you're right, your reasoning looks perfectly good to me. Although I agree with your answer, I may also be making an oversight here, so let's see if anyone else has anything to say about it. – doobdood Jan 06 '22 at 07:32

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You are right. Here is my way to see it with the following arrays that are clearly in a bijective correspondence with the binary operations.

A $5 \times 5$ array of a commutative operation has this form:

$$\begin{array}{c|ccccc} &a&b&c&d&e\\ \hline a&s_1&r_1&r_2&r_3&r_4\\ b&r_1&s_2&r_5&r_6&r_7\\ c&r_2&r_5&s_3&r_8&r_{9}\\ d&r_3&r_6&r_8&s_4&r_{10}\\ e&r_4&r_7&r_9&r_{10}&s_5\\ \end{array} $$

where we are free to place, for each of the 15 "sites" $r_1,\cdots r_{10}$ and $s_1,\cdots s_5$, one of the five elements $a,b,c,d,e$, giving rise to your solution $5^{15}$.

The solution given by the book looks to forget the diagonal choices.

Edit: I just saw a nice general answer here.

Jean Marie
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