$1\le r\le n$ and consider all $r-$element subset of $\{1,2,...,n\}$.$F(n,r)$ is the AM of the smallest elements of these subsets. $1\le r\le n$ and consider all $r-$element subset of $\{1,2,...,n\}$.$F(n,r)$ is the AM of the smallest elements of these subsets.Prove that, $F(n,r)=\dfrac{n+1}{r+1}$  
I know this is a famous problem, and it has solutions all around. One of the way it is done is bijection. But,  I cannot understand the solution. Please help. Thank you.
 A: I'm going to assume you meant The solution by Boy Soprano II.
The mean is the sum over all the $r$-sets divided by the number of $r$-sets. The number of $r$-subsets is $\binom{n}{r}$
Hence If $S$ is the sum the mean is $\frac{S}{\binom{n}{r}}$. So we must prove $S=\binom{n+1}{r+1}$ to have $\frac{S}{\binom{n}{r}}=\frac{n-1}{r-1}$.
Lets try to count $S$. To do this we separate the sum. How many subsets have minimum element $j$? The other $r-1$ elements of such a set belong to the set $\{j+1,j+2\dots n\}$ of $n-j$ elements. Therefore there are $\binom{n-j}{r-1}$ such subsets.
So we can now write $S=\sum\limits_{j=1}^n j\binom{n-j}{r-1}$.
If we could prove this sum is equal to $\binom{n+1}{r+1}$ we would be done.
We will now prove $\sum\limits_{j=1}^n j\binom{n-j}{r-1}=\binom{n-1}{r-1}$ counting two ways
We count the $r+1$ subsets of $\{1,2,3\dots n,n+1\}$ in two ways.
Clearly there are $\binom{n+1}{r+1}$ such subsets.
On the other hand how many of these sequences have $k+1$ as the second smallest element? There are $k$ possible values for the smallest element and $\binom{n-k}{r-1}$ ways to select the $r-1$ largest values. Hence there are $k\binom{n-k}{r-1}$ subsets which have $k$ as the second largest element. Adding over the possible values for $k+1$  we get: $\sum\limits_{k=1}^{n}k\binom{n-k}{r-1}=\binom{n+1}{r+1}$ as desired.
