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?

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 :
$$ \frac{\displaystyle\sum_{i=1}^{n-r+1}i \cdot \binom{n-i}{r-1}}{\displaystyle\binom{n}{r}}=\frac{\displaystyle\binom{n+1}{r+1}}{\displaystyle\binom{n}{r}} =\frac{n+1}{r+1}$$
