arithmetic mean of smallest numbers of all subsets of r elements formed out of (1,2,..n) Consider all subsets of r elements of the set $\{1,2,3,......,n\}$ where $1 \leq r \leq n$.
Each of these subsets has a smallest member. Let $F(n,r)$ denote the arithmetic mean of these smallest numbers then
$$
F(n,r)=\frac{n+1}{r+1}\;.
$$
I  got this solution on a page but couldn't understand it as how he started and what are the different steps one needs to take. Throw light on every single step.
$$\begin{align*}
\sum_{k=1}^nk\binom{n-k}{r-1}&=\sum_{k=1}^{n-(r-1)}k\binom{n-k}{r-1}\\
&=\sum_{i=1}^{n-(r-1)}\sum_{k=i}^{n-(r-1)}\binom{n-k}{r-1}\\
&=\sum_{i=1}^{n-(r-1)}\binom{n-i+1}r\\
&=\binom{n+1}{r+1}\;.
\end{align*}$$
And
$$\frac{\binom{n+1}{r+1}}{\binom{n}r}=\frac{n+1}{r+1}\;.$$

The number of ways that $k$ is the smallest of $r$ numbers from $1\dots n$ is
$$
\binom{n-k}{r-1}\tag{1}
$$
As one would expect, the total number of ways to arrange the $r$ numbers from $1\dots n$ is
$$
\sum_{k=1}^{n-r+1}\binom{n-k}{r-1}=\binom{n}{r}\tag{2}
$$
Thus, the expected smallest of $r$ numbers from $1\dots n$ times $(2)$ is
$$
\begin{align}
\sum_{k=1}^{n-r+1}\binom{n-k}{r-1}k
&=(n+1)\binom{n}{r}-\sum_{k=1}^{n-r+1}\binom{n-k}{r-1}(n-k+1)\\
&=(n+1)\binom{n}{r}-\sum_{k=1}^{n-r+1}\binom{n-k+1}{r}\,r\\
&=(n+1)\binom{n}{r}-\binom{n+1}{r+1}\,r\\
&=(n+1)\binom{n}{r}-\frac{n+1}{r+1}\binom{n}{r}\,r\\
&=\frac{n+1}{r+1}\binom{n}{r}\tag{3}
\end{align}
$$
Therefore, the expected smallest of $r$ numbers from $1\dots n$ is
$$
\frac{n+1}{r+1}\tag{4}
$$
