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If f(n)=|(n!, (n+1)!, (n+2)!),((n+1)!, (...

If `f(n)=|(n!, (n+1)!, (n+2)!),((n+1)!, (n+2)!, (n+3)!), ((n+2)!, (n+3)!, (n+4)!)|` Then the value of 1/1020[(f(100))/(f(99))]` is

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