Is there a pattern?

$\begin{aligned} 10! & = 6! \times 7 \times 8 \times 9 \times 10 \\ & = 6! \times {\color{#3D99F6} 2^4 \times 3^2} \times 5\times 7 \\ & = 6! \times {\color{#3D99F6} 144} \times 5\times 7 \\ & = 6! \times {\color{#3D99F6} 2 \times 3 \times 4 \times 6} \times 5\times 7 \\ & = 6! \times 7! \end{aligned}$

$\begin{aligned} \implies \frac {10!}{7!} & = 6! \\ \implies \frac {10!}{6!} & = 7! \end{aligned}$

Now, we have:

$\begin{aligned} 10! & = {\color{#3D99F6}6!} \times 7! \\ & = {\color{#3D99F6}5! \times 6} \times 7! \\ & = {\color{#3D99F6}3!\times 5!} \times 7! \end{aligned}$

$\begin{aligned} \implies \frac {10!}{5!} & = 3! \times 7! = 6 \times 5040 = \boxed{30240} \end{aligned}$

$\dfrac{10!}{5!}=\dfrac{1\times2\times3\times4\times5\times6\times7\times8\times9\times10}{1\times2\times3\times4\times5}=6\times7\times8\times9\times10=6\times7!=\boxed{30240}$

The previous results, which appear to follow a pattern, do so mostly by accident.

$\dfrac{10!}{7!}=\dfrac{1\times2\times3\times4\times5\times6\times7\times8\times9\times10}{1\times2\times3\times4\times5\times6\times7}=8\times9\times10=1\times2\times3\times4\times5\times6=6! =720$ .

$\dfrac{10!}{6!}=6\times7\times8\times9\times10=2^5\times3^2\times5\times7=1\times2\times3\times4\times5\times6\times7=7!=5040$ .