For each of the following pairs of functions, indicate whether the first functionof each of the following pairs has a lower, same, or higher order of growth (towithin a constant multiple) than the second function.a. n(n + 1) and 2000n2 b. 100n2 and 0.01n3c. log 2 n and ln n d. log 22 n and log 2 n2e. 2n−1 and 2n f. (n − 1)! and n!

a. n(n + 1) and 2000n2: The first function has the same order of growth as the second function. This is because n(n + 1) simplifies to n^2 + n, and when considering order of growth, we only consider the highest power of n, which is 2 in both cases.

b. 100n2 and 0.01n3: The first function has a lower order of growth than the second function. This is because the highest power of n in the first function is 2, while in the second function it is 3.

c. log 2 n and ln n: These two functions have the same order of growth. This is because the logarithm base does not affect the order of growth.

d. log 22 n and log 2 n2: The first function has a lower order of growth than the second function. This is because the first function is equivalent to log 2 n, while the second function is equivalent to 2 log 2 n.

e. 2n−1 and 2n: The first function has a lower order of growth than the second function. This is because 2n−1 is equivalent to 1/2 * 2n, and constants do not affect the order of growth.

f. (n − 1)! and n!: The first function has a lower order of growth than the second function. This is because n! is equivalent to n * (n - 1)!, and n is greater than 1 for all n > 1.