I have listed them all by brute force :
a = 0,1 : no solutions
a = 2 : b = 3,4,5,...9
c = 3 : b = 2,3,4...9
I'm wondering if there is a clever approach to this. Appreciate any help, thanks!
2026-03-28 02:07:54.1774663674
Any nice way to find number number of single digit ordered pairs $(a, b)$ such that $a!b! \gt a!+b!$
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Note that $$a!b!\gt a!+b!\iff (a!-1)(b!-1)\gt 1$$ with $a!\ge 1,b!\ge 1$.