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+3 votes

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.**

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