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.

  • 3
    $\begingroup$ You need to look at rising factorial and falling factorial and n-permutations for some of the notation used. $\endgroup$
    – Henry
    Feb 24 '12 at 0:41
  • $\begingroup$ You could restate your question as asking when is a factorial equal to the non-trivial product of two factorials as in $10! = 7! \times 6!$ and when is it the non-trivial product of three factorials as in $10! = 7! \times 5! \times 3!$. $\endgroup$
    – Henry
    Feb 24 '12 at 1:05
  • $\begingroup$ As for the notation, I was aware of the possibility someone had come up with it before me, it's not like I'm THAT good. And, you are completely right. I could've thought that one more through. It was late at night. $\endgroup$
    – Arthur
    Feb 24 '12 at 13:24
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    $\begingroup$ The notation $a!b$ has a well-established and familiar meaning: namely the product of $a!$ and $b$. $\endgroup$ Sep 15 '13 at 12:56
  • $\begingroup$ Similar question was answered at MathOverflow. $\endgroup$ Mar 30 '18 at 11:47

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-factorial 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

  • $\begingroup$ Very interesting reference, thank you! $\endgroup$
    – Jose Brox
    Feb 18 '16 at 20:43

Consider the general equation $$a_1!a_2!\cdots a_n! = b!.$$ People like Paul Erdős worked on this kind of equations. So, they are rather serious problems. Even for the simplest case where $n=2$, namely the equation $A!B!=C!$, we don't know whether there is any other non-trivial solution other than $10!=7!6!$.

Erdős proved that in the equation $A!B!=C!$ if we assume $1< A \leq B$, then for large enough values of $C$, the difference of $B$ and $C$ does not exceed $5 \log \log n$. It is clear that $5 \log \log n$ grows to infinity as $n$ increases. But in fact, this is a very slowly growing function. For instance, the first $n$ for which $5 \log \log n = 50$ has more than $9566$ digits. Caldwell proved (1994) that the only non-trivial solution with $C < 10^6$ is $10!=7!6!$. Florian Luca proved in this paper (2007) that considering the $abc$ hypothesis, we have $C-B=1$ for large enough $n$. Read also a remark (2009) on this paper if you're interested.

Suppose that $P(m)$ denotes the largest prime factor of $m$. Erdős also showed in one of his many papers that the assertion $$ P(n(n+1)) > 4 \log n $$ would imply that the general equation $(1)$ has finitely many non-trivial solutions.

The last paragraph is taken from a recent work of Hajdua, Pappa, and Szakács, who prove proved in this paper (2018) that writing $k=B-A$ for all non-trivial solutions of the equation $A!B!=C!$ different from $(A, B, C) =(6, 7, 10)$, we have $C<5k$. Further, if $k \leq 10^6$, then the only non-trivial solution is given by $(A, B, C) =(6, 7, 10)$. I presented their paper in our MathSciNet seminar at UBC and you can find the slides here.

More references:

  1. Discussion on ArtofProblemSolving
  2. Luca, Florian. "The Diophantine equation $P (x)= n!$ and a result of M. Overholt." Glasnik matematički 37.2 (2002): 269-273.
  3. Berend, Daniel, and Jørgen Harmse. "On polynomial-factorial diophantine equations." Transactions of the American Mathematical Society 358.4 (2006): 1741-1779.
  • $\begingroup$ Sir, $3! \cdot5! =6!$ doesn't this qualifies as a non-trivial solution for $A! \cdot B! =C! $ with $C<10^6$ ? Please correct me, if I'm wrong. $\endgroup$ Jul 17 '20 at 17:05

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