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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.

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You need to look at rising factorial and falling factorial and n-permutations for some of the notation used. – Henry Feb 24 '12 at 0:41
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 $10! = 7! \times 5! \times 3!$. – Henry Feb 24 '12 at 1:05
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. – Arthur Feb 24 '12 at 13:24
The notation $a!b$ has a well-established and familiar meaning: namely the product of $a!$ and $b$. – John Bentin Sep 15 '13 at 12:56
up vote 17 down vote accepted

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).

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Funny I should stumble upon half of the known solutions to an unsolved problem. Thanks for the references, though. I'll look them up. – Arthur Feb 24 '12 at 13:25

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

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Very interesting reference, thank you! – Jose Brox Feb 18 at 20:43

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