Show that arithmetic mean of numbers obtained is $\frac{n+1}{r+1}$ 
Consider the smallest number in each of the $n \choose r$ subsets (of size r) of $S = \{1, 2, \ldots, n\}$. Show that the arithmetic mean of the numbers so obtained is $\frac{n+1}{r+1}$

I've tried counting the number of times every integer appears, but it seems too complicated (for me). Is there any way this can be done without induction?
 A: It won't be complicated.
Let's talk about each number, first $1$.
The number of sets which have $1$ as smallest number is the same as selecting subsets of containig $r-1$ elements from the set $\{2,3 \dots n-1\}$.
Which is equal to : 

$$ \binom{n-1}{r-1}$$

Similarly for number $i$ to be smallest in group of $r$ elements, all the remaining $r-$ elements must be selected from the set $\{i+1,i+2 \dots n\}$.That is :

$$ \binom{n-i}{r-1}$$

The mean of these numbers will be : $$ \displaystyle\sum_{i=1}^{n-r+1}i \cdot \binom{n-i}{r-1} \over \displaystyle\binom{n}{r}$$
Now, it can be proved by induction that :
$$ \displaystyle\sum_{i=1}^{n-r+1}i \cdot \binom{n-i}{r-1}=\binom{n+1}{r+1}$$
Hence, the mean :
$$ \frac{\displaystyle\sum_{i=1}^{n-r+1}i \cdot \binom{n-i}{r-1}}{\displaystyle\binom{n}{r}}=\frac{\displaystyle\binom{n+1}{r+1}}{\displaystyle\binom{n}{r}} =\frac{n+1}{r+1}$$
A: By the tail sum formula for expectation, and the hockey stick identity,
$$E[X] = \sum_{k=1}^{n-r+1} P(X \ge k) = \frac{1}{\binom{n}{r}}\sum_{k=1}^{n-r+1} \binom{n-k+1}{r} = \frac{1}{\binom{n}{r}}\sum_{j=r}^n \binom{j}{r} = \frac{1}{\binom{n}{r}}\binom{n+1}{r+1} = \frac{n+1}{r+1}.$$
A: This is the problem of finding the expectation of a random $r$-element
subset $A$ of $\{1,2,\ldots,n\}$ (chosen uniformly). Let the elements
of $A$ be the discrete random variables $X_1<X_2<\ldots<X_r$.
So our task is to find $E(X_1)$.
Introduce new random variables:
$Y_1=X_1$, $Y_2=X_2-X_1$, $Y_3=X_3-X_2,\ldots, Y_r=X_r-X_{r-1}$, $Y_{r+1}
=n+1-X_r$.
These take positive integer values and $Y_1+\cdots+Y_{r+1}=n+1$.
Each sequence of integers which fulfils these conditions corresponds to
a unique set $A$. Therefore each such sequence occurs with the
same probability. These conditions are symmetric in $1,\ldots,r+1$
so the $Y_i$ all have the same distribution. In particular
$E(Y_1)=E(Y_2)=\cdots=E(Y_{r+1})$. So
$$(r+1)E(Y_1)=E(Y_1+Y_2+\cdots+Y_{r+1})=E(n+1)=n+1.$$
Therefore
$$E(X_1)=E(Y_1)=\frac{n+1}{r+1}.$$
As a bonus, the expectation of the $k$-th smallest element of $A$ is
$$E(X_k)=E(Y_1+\cdots+Y_k)=E(Y_1)+\cdots+E(Y_k)
=kE(Y_1)=k\frac{n+1}{r+1}.$$
