8. Since the number $k, k=1,2, \ldots, n-r+1$, is the minimum in exactly $\binom{n-k}{r-1} r$-element subsets of $\{1,2, \ldots, n\}$, it follows that
$$f(n, r)=\frac{1}{\binom{n}{r}} \sum_{k=1}^{n-r+1} k\binom{n-k}{r-1}$$
To calculate the sum in the above expression, using the equality $\binom{r+j}{j}=$ $\sum_{i=0}^{j}\binom{r+i-1}{r-1}$, we note that
$$\begin{aligned} \sum_{k=1}^{n-r+1} k\binom{n-k}{r-1} & =\sum_{j=0}^{n-r}\left(\sum_{i=0}^{j}\binom{r+i-1}{r-1}\right) \\ & =\sum_{j=0}^{n-r}\binom{r+j}{r}=\binom{n+1}{r+1}=\frac{n+1}{r+1}\binom{n}{r} . \end{aligned}$$
Therefore $f(n, r)=(n+1) /(r+1)$.