9
8
7
6
5

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$\large n!$ will have 3 more zeros in its decimal expansion than $\large (n-1)!$ exactly when $\large n$ is divisible by $5^{3}=125.$ There are seven 3 digit multiples of 125 since $8.125=1000$ . But we can't include 625 because there is a difference of 4 zeros between $624! and 625!$ because $625 = 5^{4}$ and hence only values are $\large 125!,250!,375!,500!,750!,875!$ . Total values $\huge 6$