Solution

The correct answer is option 1.

Explanation: 

Here are some common Big O notations from the smallest to the largest complexity:

• O(1): Constant-time complexity. The running time is independent of the input size, it always takes the same time to compute.
• O(log n): Logarithmic time complexity. The running time grows logarithmically in proportion to the input size.
• O(n): Linear time complexity. The running time grows linearly with the size of the input.
• O(n log n): Log-linear or Linearithmic time complexity. Running time grows linearly and in the order of the logarithm of the size of the input.
• O(n^2): Quadratic time complexity. The running time is proportional to the square of the size of the input data.
• O(n^3): Cubic time complexity. The running time is proportional to the cube of the size of the input data.
• O(2^n): Exponential time complexity. The resources needed for algorithm execution double with each addition to the input data set.
• O(n!): Factorial time complexity. The running time grows in proportion to the factorial of the input size. This is typical of algorithms that solve problems by generating all possible combinations.

So when you say "all types of asymptotic complexity increasing order", we could be talking about a set of functions like:

• f1(n) = 1 (O(1))
• f2(n) = log n (O(log n))
• f3(n) = n (O(n))
• f4(n) = n log n (O(n log n))
• f5(n) = n^2 (O(n^2))
• f6(n) = n^3 (O(n^3))
• f7(n) = 2^n (O(2^n))
• f8(n) = n! (O(n!))

So, according to above explanation:

•  f1(n) = 2ⁿ
•  f2(n) = n^3/2
•  f3(n) = n log n 
•  f4(n) = n^{log n}

The functions in ascending order: n log n  <  n^3/2 < n^{log n}   <   2ⁿ