how to solve factorial involving multiplication

Question

I am trying to solve this question but not able to find any helpful material. It involves factorial with multiplications,

$$\frac{8!}{5!}\cdot \frac{7!}{7!10!}$$

I tried crossing 8 and 5 and 7 with 7 but it's not giving me right answer

Answer 1

$\dfrac{8!}{5!}\cdot \dfrac{7!}{7!10!}=\dfrac{8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}{5\cdot4\cdot3\cdot2\cdot1}\cdot\dfrac{7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}{7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1\cdot10!}\\=8\cdot7\cdot6\cdot\dfrac{1}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}\\=\dfrac{8\cdot7\cdot6}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}=\dfrac{1}{10\cdot9\cdot5\cdot4\cdot3\cdot2\cdot1}\\=\dfrac{1}{10800}$.

You can also think of it like the following:

$\dfrac{8!}{5!}\cdot \dfrac{7!}{7!10!}=\dfrac{8\cdot7!}{5!}\cdot\dfrac{7\cdot6\cdot5!}{7!10!}\\=\dfrac{8\cdot7\cdot6}{10!}=\dfrac{8\cdot7\cdot6}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}\\=\dfrac{1}{10\cdot9\cdot5\cdot4\cdot3\cdot2\cdot1}\\=\dfrac{1}{10800}$

Answer 2

my steps: $ \dfrac{8!}{5!}\cdot\dfrac{7!}{7!\cdot10!} $ 7! and 7! cancel out: $\dfrac{8!}{5!} \cdot \dfrac{1}{10!}$

next: $ \dfrac{8!}{5!} \cdot \dfrac{1}{10 \cdot 9 \cdot 8!} $ Here, 8! cross-cancel : $\dfrac{1}{5!} \cdot \dfrac{1}{10 \cdot 9} $

next: $ \dfrac{1}{10 \cdot 9 \cdot 5!}$

which is $\dfrac{1}{ 10 \cdot 9 \cdot 5 \cdot4 \cdot3 \cdot2\cdot1}$ which ended up being $ 9.26 * 10^{-5} $ or $\dfrac{1}{10800}$

Answer 3

$\dfrac{8!}{5!}\cdot \dfrac{7!}{7!10!} = \dfrac{8!}{5!\cdot 8! \cdot 9\cdot10} = \dfrac{1}{120 \cdot90} = \dfrac{1}{10800} $.