I try to prove this bt am confuse why divided by (p-1) .I know it is permutations question bt which case apply here

Question

am confuse why here divide by(p-1) I know it is permutations case bt which case is apply here

Answer 1

If you want to understand the why of the numerator, see Marcus M's answer to the question

$\qquad$What is the number of invertible $n\times n$ matrices in $\operatorname{GL}_n(F)$?

As to your question about why there is a denominator of $p-1$, it can be explained as follows . . .

Fix a positive integer $n$.

Let $X$ be the set of $n\times n$ matrices with entries in $Z_p$ which are non-singular, mod $p$.

For each $d \in Z_p\setminus\{0\}$, let $X_d = \{A \in X\mid (\det(A)\;\text{mod}\; p) = d\}$.

Claim: The sets $S_d$ all have the same cardinality.

To prove it, let $d,e \in Z_p\setminus \{0\}$.

Define $f:X_d\to X_e$ as follows . . .

For each matrix $A \in X_d$, let $f(A)$ be the matrix whose first row is the first row of $A$ scaled, mod $p$, by ${\large{\frac{e}{d}}}$, and all of whose other entries are the same as the corresponding entries in $A$.

It's easily seen that $f$ is injective.

Thus, $|X_d| \le |X_e|$.

Using the same reasoning, we can get an injective function from $X_e$ to $X_d$, hence $|X_e| \le |X_d|$.

Thus, $|X_d|=|X_e|$, as claimed.

Since $|Z_p\setminus \{0\}|=p-1$, and all the sets $X_d$ have the same cardinality, it follows that $$|X_d|=\frac{|X|}{p-1}$$ which explains the reason for dividing by $p-1$.