1.3.11 problem 40

(a)

Counting strictly increasing sequences of $k$ numbers amounts to counting the number of ways to select $k$ elements out of the $n$, since for any such selection, there is exactly one increasing ordering. Thus, the answer is $$\frac{\left(\frac{n}{k}\right)}{n^k}$$

(b)

The problem can be thought of sampling with replacement where order doesn’t matter, since there is only one non decreasing ordering of a given sequence of $k$ numbers. Thus, the answer is $$\frac{\left(\frac{n-1+k}{k}\right)}{n^k}$$