Considering \(s\) and \(t\) are integers greater than 1 and factors of \(n\),

\(2 \le s,t \le n\)

\(n^{st}\) can be written as \((n^s)^t = n^s*n^s*n^s*......\)t times

I - \(s^t = s*s*s*s*......\)t times

Since there are \(t\) terms of both \(n^s\) and \(s\), there is one to one correspondence \((n^s \rightarrow s)\)

If \(f\) is a factor of \(p\), then \(f\) will also be a factor of \(p^q\) where \(q\) is any integer greater than 0.

Since both \(n^{st}\) and \(s^t\) have \(t\) terms each and based on above property, \(s\) will be a factor of \(n^s\), overall \(s^t\) will be a factor of \(n^{st}\)

II - \((st)^2\)

Since minimum value of both \(s\) and \(t\) is 2, the minimum value of \(n^{st}\) will be \(n^{2*2} = n^4 = n*n*n*n\)

Now \((st)^2 = s*s*t*t\), we can clearly see that there is one to one correspondence here.

\([n \rightarrow s] | [n \rightarrow s] | [n \rightarrow t] | [n \rightarrow t]\).

Therefore, \((st)^2\) is a factor of \(n^{st}\)

III - \((s+t)\)

Take the min value of \(s\) and \(t\), then \(s+t = 2+2 = 4\) and \(n^{st}\) will be \(n^{2*2} = n^4\)

Now all we have to do is check if 4 is a factor of \(n^4\).

If we take \(n = 5\), then \(4\) will be not be a factor. But if we take \(n = 4\) then \(4\) will be a factor.

Hence, \(s+t\) is not a factor in all cases.

Hence, only I and II are factors of \(n^{st}\)

Hence, Answer is E