Recently, I have been involved in creating interesting questions. This is an example of one of the questions I have created.
Question 1: Find integers $ a, b $ and $ c $, where $ a, b, c > 2 $ such that $ a! = b! \cdot c! $.
Question 2: Find integers $ a, b, c $ and $ d $, where $ a, b, c, d > 2 $ such that $ a! = b! \cdot c! \cdot d! $.
The solution for these problems are not hard to find. A hint can be found below (don't look at the hint before thinking it through):
This question is related to binomial coefficients. To solve it, you could look at the Pascal triangle.
However, I ran into a problem when extending to $ 5 $ integers, i.e. to find $ a, b, c, d $ and $ e $, where $ a, b, c, d, e > 2 $ such that $ a! = b! \cdot c! \cdot d! \cdot e! $.
I wrote a code to check for small integers and found none. For larger integers, my computer took too long to evaluate (it could be that my code is not optimized, admittedly I'm not a good coder). So, I am just curious is there actually a solution for $ 5 $ integers?
In general, for any integers $c,d,e > 2$, you can set $b = c! \cdot d! \cdot e! - 1$ and $a = c! \cdot d! \cdot e!$.
Then, $a! = (a-1)! \cdot a = (c! \cdot d! \cdot e!-1)! \cdot (c! \cdot d! \cdot e!) = b! \cdot c! \cdot d! \cdot e!$.
These aren't necessarily all the solutions, but this does show that there are infinitely many solutions.