Consider all subsets of r elements of the set $\{1,2,3,......,n\}$ where $1 \leq r \leq n$. Each of these subsets has a smallest member. Let $F(n,r)$ denote the arithmetic mean of these smallest numbers then $$ F(n,r)=\frac{n+1}{r+1}\;. $$ I got this solution on a page but couldn't understand it as how he started and what are the different steps one needs to take. Throw light on every single step.

$$\begin{align*} \sum_{k=1}^nk\binom{n-k}{r-1}&=\sum_{k=1}^{n-(r-1)}k\binom{n-k}{r-1}\\ &=\sum_{i=1}^{n-(r-1)}\sum_{k=i}^{n-(r-1)}\binom{n-k}{r-1}\\ &=\sum_{i=1}^{n-(r-1)}\binom{n-i+1}r\\ &=\binom{n+1}{r+1}\;. \end{align*}$$



  • $\begingroup$ You can typeset it beautifully using MathJax (this is a link to a turotial). $\endgroup$ – String Apr 27 '16 at 9:11
  • $\begingroup$ @String I don't have enough knowledge of Matjax $\endgroup$ – satyajeet jha Apr 27 '16 at 9:12
  • $\begingroup$ @string it is (n+1)/(r+1) not what you wrote $\endgroup$ – satyajeet jha Apr 27 '16 at 9:17
  • $\begingroup$ @string i edited it completely so its clear now ,i think $\endgroup$ – satyajeet jha Apr 27 '16 at 9:23

The number of subsets having $k$ as its smallest element will be $$ \binom{n-k}{r-1} $$ since the remaining $r-1$ elements in such a subset must be chosen from the remaining $n-k$ larger elements. But we cannot choose $r-1$ elements unless we have $n-k\geq r-1$ elements to choose from. Thus it follows by rearranging that $1\leq k\leq n-(r-1)$. So the sum of the smallest elements $k$ counted by multiplicity $\binom{n-k}{r-1}$ must be: $$ \sum_{k=1}^{n-(r-1)}k\binom{n-k}{r-1} $$

Next step uses that $$ \sum_{k=1}^m k\cdot f(k)=\sum_{i=1}^m\sum_{j=i}^m f(j) $$ since in the second double sum a given value $k$ of $j$ is only iterated through if $i\leq k$ which happens exactly $k$ times, namely for $i=1,2,...,k$.

For the final two steps the principle is $$ \sum_{s=a}^b\binom{s}{a}=\binom{a}{a}+\binom{a+1}{a}+...+\binom{b}{a}=\binom{b+1}{a+1} $$ which is true since choosing $a+1$ elements from a set containing $b+1$ elements can be done by choosing the smallest element first and then choosing the remaining $a$ elements from the remaining $s\in[a,b]$ larger elements.

The first time it is used where $s=n-k$ ranges from $a=r-1$ to $b=n-i$ and the second time it is used where $s=n-i+1$ ranges from $a=r$ to $b=n$. To be explicit: $$ \sum_{k=i}^{n-(r-1)}\binom{n-k}{r-1}=\sum_{s=r-1}^{n-i}\binom{s}{r-1}=\binom{n-i+1}{r} $$ and $$ \sum_{i=1}^{n-(r-1)}\binom{n-i+1}{r}=\sum_{s=r}^n\binom{s}{r}=\binom{n+1}{r+1} $$

So now we have the sum of all the smallest elements. Then we just need to divide by the number of them, namely $\binom nr$ and so the average smallest element will be: $$ \mu=\frac{\binom{n+1}{r+1}}{\binom{n}{r}}=\frac{n+1}{r+1} $$

  • $\begingroup$ please explain the remaining part after second = to in solution as how it has been done. $\endgroup$ – satyajeet jha Apr 27 '16 at 9:35
  • $\begingroup$ @satya: I am working on it ... $\endgroup$ – String Apr 27 '16 at 9:35

The number of ways that $k$ is the smallest of $r$ numbers from $1\dots n$ is $$ \binom{n-k}{r-1}\tag{1} $$ As one would expect, the total number of ways to arrange the $r$ numbers from $1\dots n$ is $$ \sum_{k=1}^{n-r+1}\binom{n-k}{r-1}=\binom{n}{r}\tag{2} $$ Thus, the expected smallest of $r$ numbers from $1\dots n$ times $(2)$ is $$ \begin{align} \sum_{k=1}^{n-r+1}\binom{n-k}{r-1}k &=(n+1)\binom{n}{r}-\sum_{k=1}^{n-r+1}\binom{n-k}{r-1}(n-k+1)\\ &=(n+1)\binom{n}{r}-\sum_{k=1}^{n-r+1}\binom{n-k+1}{r}\,r\\ &=(n+1)\binom{n}{r}-\binom{n+1}{r+1}\,r\\ &=(n+1)\binom{n}{r}-\frac{n+1}{r+1}\binom{n}{r}\,r\\ &=\frac{n+1}{r+1}\binom{n}{r}\tag{3} \end{align} $$ Therefore, the expected smallest of $r$ numbers from $1\dots n$ is $$ \frac{n+1}{r+1}\tag{4} $$


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