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I have to find $a,b,c \in \mathbb{N}$ such that-$a!\times b!\times c!=d!$

Answer given in my book is $3!\times5!\times7!=10!$(But it is written that other answers are also possible).

What is a systematic way to find the answer without using a calculator or computer?(I mean without brute force calculation)

Thanks for any help!!


marked as duplicate by Jyrki Lahtonen Mar 13 '16 at 7:59

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  • $\begingroup$ One set of solutions is a,b,c and d are in the set {0,1}. $\endgroup$ – user145600 Mar 13 '16 at 7:36
  • $\begingroup$ related : On the factorial equations $A! B! =C!$ and $A!B!C! = D!$ $\endgroup$ – mathlove Mar 13 '16 at 7:37
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    $\begingroup$ There are many uninteresting answers, such as $12!=11!\cdot 12=11!\cdot 2!\cdot 3!$. $\endgroup$ – André Nicolas Mar 13 '16 at 7:41
  • $\begingroup$ @AndréNicolas Your answer provides one way of approaching the solution....But you are fortunate that 12 can be expressed as a product of two factorials...not all numbers can be expressed as a product of two factorials:-)... $\endgroup$ – tatan Mar 13 '16 at 7:44
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    $\begingroup$ @tatan The point is we can always take $a=b!c!-1,\,d=b!c!$ for a trivial solution. $\endgroup$ – J.G. Mar 13 '16 at 7:53