All possible subsets containing $r$ elements from the set$\{1,2,3,...,n\}$ are formed where $(1\leq r\leq n)$. What is the arithmetic mean of the smallest elements of these subsets.

My Attempt

The number of such subsets is clearly $\binom{n}{r}$. The number of subsets with $k$ as its smallest element is $\binom{n-k}{r-1}$. So sum of all the smallest elements$$\sum_{k=1}^{n-r+1}k\binom{n-k}{r-1}$$

=Coefficient of $x^{r-1}$in expansion$$\sum_{k=1}^{n-r+1}k(1+x)^{n-k}$$ =coefficient of $x^{r-1}$ in expansion $$\left\{\frac{(1+x)^{n+1}-(1+x)^r}{x^2}-\frac{(n-r+1)(1+x)^{r-1}}{x}\right\}=\binom{n+1}{r+1}$$

So required AM$$=\frac{n+1}{r+1}$$

Is there a combinatorical argument to this.


We count the number of $r+1$ element subsets of $\{0,1,...,n\}$. On one hand, this is $\binom{n+1}{r+1}$. On the other hand, for each $1 \leq k \leq n$, consider the number of $r+1$ element subsets of $\{0,1,...,n\}$ with second smallest element $k$. This can be counted by first taking a $r$ element subset of $\{1,...,n\}$ with smallest element $k$, then having $k$ ways to choose the last element from $\{0,...,k-1\}$. Therefore the number is equal to the sum of the smallest element of $r$ element subsets of $\{1,...,n\}$ with smallest element $k$. Summing over $k$, the sum of smallest elements of $r$ element subsets of $\{1,...,n\}$ is the number of $r+1$ element subsets of $\{0,1,...,n\}$, i.e. $\binom{n+1}{r+1}$. We then get the A.M. to be $\frac{n+1}{r+1}$ upon dividing by $\binom{n}{r}$.


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