Find all such positive integers $x(>1), y(>1)$ and $z$ so that $x! \cdot y! =z!$ holds.

Question

Upon inspection, I found out that: $3!\cdot5! =6! $ and $6!\cdot7! =10!.$ (couldn't dare to search for more such triplets)

How to check whether there exist more such triplets or not? Please show me a proper way to handle such problems.

Answer 1

Your $3!\cdot 5! =6!$ is an example of the case for general $n$:

$$n! \cdot (n!-1)! = (n!)!$$

and when $n=4$ this gives $4! \cdot 23! = 24!$

Answer 2

For example, you can observe that

$$(n!-1)!.(n!)=(n!)!$$

If $ n=3$, it gives

$$5!.3!=6!$$ for $ n=4 $, we get

$$23!.4!=24!$$ and so on.