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?
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 : $$ \frac{\displaystyle\sum_{i=1}^{n-r+1}i \cdot \binom{n-i}{r-1}}{\displaystyle\binom{n}{r}}=\frac{\displaystyle\binom{n+1}{r+1}}{\displaystyle\binom{n}{r}} =\frac{n+1}{r+1}$$
By the tail sum formula for expectation, and the hockey stick identity, $$E[X] = \sum_{k=1}^{n-r+1} P(X \ge k) = \frac{1}{\binom{n}{r}}\sum_{k=1}^{n-r+1} \binom{n-k+1}{r} = \frac{1}{\binom{n}{r}}\sum_{j=r}^n \binom{j}{r} = \frac{1}{\binom{n}{r}}\binom{n+1}{r+1} = \frac{n+1}{r+1}.$$
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 $$E(X_k)=E(Y_1+\cdots+Y_k)=E(Y_1)+\cdots+E(Y_k) =kE(Y_1)=k\frac{n+1}{r+1}.$$
