Let  $f(n,r)$  be arithmetic mean of minimum element of all subsets of  $\{1,2,3,\mathrm{.....},n\}$  which contains exactly r elements then

The number of subsets  having k as its smallest element will be ${ }^{n-k}{C}_{r-1}$, since  a subset must be chosen from the remaining n-k larger elements .

So sum of the smallest elements  k counted 

$\sum _{k=1}^{n-(r-1)}k\left({ }^{n-k}{C}_{r-1}\right)={ }^{n+1}{C}_{r+1}$

$AM=\frac{{ }^{n+1}{C}_{r+1}}{{ }^n{C}_r}=\frac{n+1}{r+1}$

$f(n,r)=\frac{n+1}{r+1}$