Question

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Answer 1

$\displaystyle \mathtt{A_x^5=5A_x^4}\\ \\ \mathtt{ \left\{\begin{array}{ccc}\mathtt{x \in N}\\\mathtt{x \geq 4}\\\mathtt{x \geq 5}\end{array\Rightarrow x \in \{5,~6,~7,~8,...\}=D}\right}\\ \\ \mathtt{\mathbf{A_n^k= \frac{n!}{(n-k)!} }\Rightarrow \left[\begin{array}{ccc}\mathtt{A_x^5= \frac{x!}{(x-5)!} }\\\mathtt{A_x^4= \frac{x!}{(x-4)!} }\end{array}\right}$

$\displaystyle \mathtt{ \frac{x!}{(x-5)!} =5 \cdot \frac{x!}{(x-4)!} }\\ \\ \mathtt{ \frac{(x-5)! (x-4) (x-3) (x-2) (x-1) x}{(x-5)!} = 5 \cdot \frac{(x-4)!(x-3)(x-2)(x-1)x}{(x-4)!} } \\ \\ \mathtt{x-4=5 \Rightarrow x=5+4 \Rightarrow x=9 \in D}\\ \\ \mathtt{S=\{9\}}$