Arrange the following functions in decreasing asymptotic order:

f(n)= 3^{2^n}

g(n)= n!n^{3}

(a) f(n),g(n)

(b) g(n),f(n)

Apply log on both functions

log(3^{2^n}) = 2^{n} log3 = O(2^{n})

log(n!n^{3}) = log n! + 3logn = O(nlogn) [ we know log n! = O(nlogn) ]

O(2^{n}) > O(nlogn)

So, option (a) is correct.