Correct Option (C) 2 Only
Explanation:
(1) n!= Ɵ((n+1)!) implies that { (n! <= C1 . (n+1)! ) and (n! >= C2 . (n+1)! ) }
Where C1 and C2 are Some Constant. [By Definition of Ɵ- Notation ]
since (n+1)! =(n+1) * n!
So (n! <= C1 . (n+1) * n! ) and (n! >= C2 . (n+1) * n! )
( 1 <= C1 . (n+1) ) and (1 >= C1 . (n+1) )
( TRUE and FALSE ) = FALSE
So (1) Statement is FALSE.
(2) log n base 4 =Θ (log n base 2)
(log n base 4) can be written as ((log n base 2) * (log 4 base 2) )
= ((log n base 2) * 2 )
So (2) statement is TRUE.
(3) Sqrt(log n) = O (log log n)
= Sqrt (log n) < = C . (log log n)
= (1/2) . log (log n) < = C .{ log log log n } [ By Taking log of both sides]
= FALSE
So (3) Statement is False.