In general, n is should be a variable unless it is explicitly mentioned that n is a constant.
If it is mentioned that n is a constant then n!, logn, (logn)!, log(n!), & lognlogn all of them would be constant for a given base of log.
Otherwise none of them would be.
Also if you are supposing n to be a constant then Logn is a smaller constant than n( unless the base of log is smaller than 1),
& n! is a constant bigger than n.
So in F1 we are taking factorial of a SMALLER CONSTANT & in F2 we are taking log of a BIGGER CONSTANT, so you can not argue like that.
It will depend on the difference between the two constants logn & n!.
------------------------------------------------------------------------------------------------
Answer to this question should be:
F2 < F1 < F3 for large values of n.
F1 = (logn)!
F2 = log(n!) = log(1 x 2 x 3 x … x (n – 1) x n) = log1 + log2 + log3 +…+ log(n – 1) + logn.
It can be observed that F2 can’t be more than nlogn.
let us consider F2 = nlogn, through out this analysis & at the end we will show that our final results would not be affected by this approximation.
F3 = lognlogn
Consider 2 is the base of logarithm, for F1, F2 & F3.(Any base >= 1 will work perfectly)
Consider n = 2y
then F1 = (log(2y))! = y!
F2 = (2y)log(2y) = y(2y)
F3 = (log(2y))log(2^y) = yy
Clearly y! is less than yy, it means F1 < F3.
Now to compare F2 with F1 and F3, let us consider n = 210, then:
F1 = 10!,
F2 = 10 x (210) ∽ 10 x 1000 = 104
F3 = 1010,
and for n = 2100
F1 = 100!
F2 = 100x(2100) ∼ 100 x (100010) = 10(30 + 2) = 1032.
F3 = 100100 = 10200
So it is clear that F2 < F3.
Now, How to compare F1 & F2 ?
for n = 210,
F1 = 10! = 1 x 2 x 3 x … x 10.
Writing 1 x 2 x 3 x … x 10 a little bit differently,
10! = 1 x 2 x (2 + 1) x 22 x (22 + 1) x (22 + 2) x (22 + 3) x 23 x (23 + 1) x 10
The LEADING TERM of the above expression will be: 10 x 2(0 + 1 + 1 + 2 + 2 + 2 + 2 + 3 + 3)= 10x216
F2 = 10 x 210
Similarly for n = 2100.
So clearly F2 < F1,
Now we have, F1 < F3, F2 < F3 & F2 < F1
So we can conclude that F2 < F1 < F3.
The only considerable approximation we have used in this analysis is F2 = log(n!) = nlogn
(although we also approximated 2^10 to 10^3 while comparing F2 & F3 but it is not going to affect the results since the difference between F2 & F3 is too big, even it is greater than F2 itself)
Even after giving F2 an extra edge over F1 & F3 by making it equal to its upper bound, it is still smaller than F1 & F3.
This means that our approximation for F2 has not affected our final results.
So the answer would be F2 < F1 < F3 for large values of n.( approximately for n >= 25)
Analysis of growth rates of F1, F2, & F3.
As n goes from 210 to 2100 ,
F1 goes from 10! to 100!, so change in F1 = 10!((100!/10!) - 1)
F2 goes from 10x210 to 100x2100 so change in F2 = 10 x 210 (10 x 290 - 1)
F3 goes from 1010 to 10200 so change in F3 = 1010 (10190 - 1)
So roughly we can see here that growth rate is also F2 < F1 < F3, so growth rate of F1, F2 & F3 are also not going to affect our answer.
For more precise analysis of growth rate, after approximating n! using Stirling's Formula, derivatives of F1, F2, & F3 can be used.