Number of Binary Operations(Compositions) with a specific neutral element Let $E$ be a finite set of $n$ elements.

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*If $a\in \mathbb{E}$, for how many compositions on $\mathbb{E}$ is is $a$ the neutral element.


*Of these how many are commutative?
 A: Any composition on $\mathbb{E}$ is a function $f:(\mathbb{E}X\mathbb{E})\Rightarrow \mathbb{E}$, where |$\mathbb{E}X\mathbb{E}$|$=n^2$. If a composition is to have $a$  as a neutral element all the ordered pairs in  $(\mathbb{E}X\mathbb{E})$ having $a$ as a member must point to the other member of the pair ,e.g.,$(a,a_i)\Rightarrow a_i$. There are $2n-1$ ordered pairs in  $(\mathbb{E}X\mathbb{E})$ with $a$ as member. Since those ordered pairs images' are already fixed we are only concerned with the rest of the ordered pairs. There are $n^2-2n+1$ ordered pairs not containing $a$ as a member.
We can therefore think of the compositions as from a set of cardinality $n^2-2n+1$ into $\mathbb{E}$ of cardinality $n$;we think of the domain as having shrunk to $n^2-2n+1$ . Therefore, the total number of compositions w/ $a$ as a neutral element is $n^{n^2-2n+1}$. 
b) We can apply the procedure for finding total number of binary operations on our now shrunken domain, see How to proof that more than half binary algebraic operations on a finite set are non-commutative?. 
1) There are $n-1$ ordered pairs where both elements are the same since $(a,a)$ is not included. 
2) Since any ordered pair having $a$ as a member is excluded we can select $\begin{pmatrix}
        n-1 \\
        2  \\
        \end{pmatrix}$ ordered pairs where the elements are not repeated. 
now we the domain as shrunken to  $\begin{pmatrix}
        n-1 \\
        2  \\
        \end{pmatrix}$ + $n-1$. 
Therefore, we can create $n^{(n^2-n)/2}$ binary operations w/ $a$ as the neutral element that are commutative. 
Q.E.D.
