Let n ≥ 8. Give a big-Θ estimate for the number of circuits of length 8 in Kn. Θ(n) Θ(n8) Θ(8) Θ(8n) Θ(8n)

A complete graph ${\displaystyle Kn(n\Rightarrow 1)}$ is a simple graph with n vertices and an edge between every pair of vertices.
Definition permutation (order is important): ${\displaystyle nPr=n\frac{!}{n-r}!}$
Definition combination (order is not important)| ${\displaystyle nCr=\left(nr\right)=n\frac{!}{r!(n-r)!}}$ with ${\displaystyle n\ne n\cdot (n-1)\cdot \dots \cdot 2\cdot 1}$
Solution Let n be the number of vertices in Kn
When $n<8$, the graph contains less than 8 vertices and thus the graph can then not contain a circuit of length 8. 0 circuits of length 8 when $n<8$
When ${\displaystyle n\Rightarrow 8}$, any circuit of length 8 is a sequence of 8 vertices (as the last vertex in the sequence will be the first circuit again). The order of the vertices matters (as a different order will lead to the same sequence), thus we use the definition of a permutation. ${\displaystyle nPs=n\frac{!}{n-8}\ne n(n-1)(n-2)\dots (n-7)}$
Every circuit will be counted twice in the $n(n-1)(n-2)\cdots (n-7)$ sequences, as the opposite order of the sequence of vertices will lead to the same vertex, thus there are ${\displaystyle n(n-1)(n-2)\dots \frac{n-7}{2}}$ circuits of length 8.
Since $n(n-1)(n-2)\cdots (n-7)$ is a polynominal function of degree 8, ${\displaystyle n(n-1)(n-2)\dots (n-7)is\Theta \left(n^8\right)}$