jmc

Since $n!$ contains the product $2\cdot 5\cdot 10=100$ whenever $n\ge 10$, it suffices to determine the tens digit of $$7!+8!+9!=7!(1+8+8\cdot 9)=5040(1+8+72)=5040\cdot 81.$$This is the same as the units digit of $4\cdot 1$, which is $\boxed{4}$.