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see the difference . The 3! and 4! differ by 21. as the value get large they differ by a large amount . ex 11! and 10!. so we can't say 1 well we can't say 4 also as logn> less than loglogn. in option 2. log n base 4 = logn base2 *1/2 so complexity will be same.
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(n+1)!  = (n+1)* (n)*(n-1)*........1  =  (n+1)*(n!)   > n!

Hence   n! =  O((n+1)!) => option (1) is wrong.

log4n  = (1/2)* log2n, So we can write log2n as follows:

(1/2)* log2n  ≤ (1/2)* log2n ≤ log2n   => (1/2)* log2n  = ⊝(log2n)  => log4n = ⊝(log2n)  Hence (!!) is correct.

√logn ≥ loglogn => loglogn  = O(√logn)   => (!!!) is wrong.

Hence only (!!) is correct. Hence option (C) is correct.

Answer:

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