# On the factorial equations $A! B! =C!$ and $A!B!C! = D!$

I was playing around with hypergeometric probabilities when I wound myself calculating the binomial coefficient $\binom{10}{3}$. I used the definition, and calculating in my head, I simplified to this expression before actually calculating anything
$$\frac {8\cdot9\cdot10}{2\cdot3} = 120$$
And then it hit me that $8\cdot9\cdot10 = 6!$ and I started thinking about something I feel like calling generalized factorials, which is just the product of a number of successive naturals, like this
$$a!b = \prod_{n=b}^an = \frac{a!}{(b-1)!},\quad a, b \in \mathbb{Z}^+, \quad a\ge b$$
so that $a! = a!1$ (the notation was invented just now, and inspired by the $nCr$-notation for binomial coefficients). Now, apart from the trivial examples $(n!)!(n!) = n!$ and $a!1 = a!2 = a!$, when is the generalized factorial a factorial number? When is it the product of two (non-trivial) factorial numbers? As seen above, $10!8$ is both.

#### Solutions Collecting From Web of "On the factorial equations $A! B! =C!$ and $A!B!C! = D!$"

See Chris Caldwell, The diophantine equation $A!B!=C!$, J. Recreational Math. 26 (1994) 128-133. $9!=7!3!3!2!$, $10!=7!6!=7!5!3!$, and $16!=14!5!2!$ were the only known non-trivial examples of a factorial as a product of factorials as of the 3rd edition of Guy, Unsolved Problems In Number Theory (Problem B23).

Immediately we see that any $n$ that’s a power of 2 should work. e.g.

$4! = 3!2!2!$, $8! = 7!2!2!2!$, $16! = 15!2!2!2!2!$

Similarly, by combining whatever small factorials we want, we can take $n!$ and $(n-1)!$ to have that ratio. For instance, suppose we wanted to use $3!5!7! = 3628800$ somewhere; we can then make $3628800! = 3628799!7!5!3!$. You can obviously generate an infinite number of solutions this way.

However, the instances of $n!/(n-1)!$ are probably what Caldwell (see other answer) referred to as “trivial”. The non-trivial instances of $n!/(n-k)!$ would be given by

$n(n-1)(n-2) … (n-k) = A_1!A_2!…A_i!$ which — in general — I know no way of tackling the general case, but the case where $i=1$ (that is, $n!=(n-k)!A!$) can be attacked:

$n(n-1)(n-2) … (n-k) = P(n) = A!$

Which is referred to as a polynomial-diophantine equation. This is still largely open, but has some interesting results proven about it — namely, that the set of solutions is finite for any polynomial P. An extensive treatment of the matter is given in http://www.ams.org/journals/tran/2006-358-04/S0002-9947-05-03780-3/S0002-9947-05-03780-3.pdf