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
$$\frac{\binom{n+1}{r+1}}{\binom{n}r}=\frac{n+1}{r+1}\;.$$
 A: The number of subsets having $k$ as its smallest element will be
$$
\binom{n-k}{r-1}
$$
since the remaining $r-1$ elements in such a subset must be chosen from the remaining $n-k$ larger elements. But we cannot choose $r-1$ elements unless we have $n-k\geq r-1$ elements to choose from. Thus it follows by rearranging that $1\leq k\leq n-(r-1)$. So the sum of the smallest elements $k$ counted by multiplicity $\binom{n-k}{r-1}$ must be:
$$
\sum_{k=1}^{n-(r-1)}k\binom{n-k}{r-1}
$$

Next step uses that
$$
\sum_{k=1}^m k\cdot f(k)=\sum_{i=1}^m\sum_{j=i}^m f(j)
$$
since in the second double sum a given value $k$ of $j$ is only iterated through if $i\leq k$ which happens exactly $k$ times, namely for $i=1,2,...,k$.

For the final two steps the principle is
$$
\sum_{s=a}^b\binom{s}{a}=\binom{a}{a}+\binom{a+1}{a}+...+\binom{b}{a}=\binom{b+1}{a+1}
$$
which is true since choosing $a+1$ elements from a set containing $b+1$ elements can be done by choosing the smallest element first and then choosing the remaining $a$ elements from the remaining $s\in[a,b]$ larger elements.

The first time it is used where $s=n-k$ ranges from $a=r-1$ to $b=n-i$ and the second time it is used where $s=n-i+1$ ranges from $a=r$ to $b=n$. To be explicit:
$$
\sum_{k=i}^{n-(r-1)}\binom{n-k}{r-1}=\sum_{s=r-1}^{n-i}\binom{s}{r-1}=\binom{n-i+1}{r}
$$
and
$$
\sum_{i=1}^{n-(r-1)}\binom{n-i+1}{r}=\sum_{s=r}^n\binom{s}{r}=\binom{n+1}{r+1}
$$

So now we have the sum of all the smallest elements. Then we just need to divide by the number of them, namely $\binom nr$ and so the average smallest element will be:
$$
\mu=\frac{\binom{n+1}{r+1}}{\binom{n}{r}}=\frac{n+1}{r+1}
$$
A: 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
$$
\frac{n+1}{r+1}\tag{4}
$$
