I was trying to simply an expression in an exercise related to randomized algorithms. Here is the expression which I have obtained at the end.

$$ \displaystyle\frac{\displaystyle\sum_{i=1}^{n+k-1} i \binom{n-i}{k-1}}{ \displaystyle{n \choose k}}$$

Is there any way to simplify the numerator so that the whole expression simplifies into a nice closed formula? A combinatorial approach would be greatly appreciated.

  • $\begingroup$ A quick (but perhaps not too helpful) simplification of the numerator, eliminating the zero terms: $$\sum_{i=1}^{n-k+1} i\binom{n-i}{k-1}$$ $\endgroup$ – Zubin Mukerjee Jan 6 '15 at 6:02
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    $\begingroup$ @ZubinMukerjee: That is not equivalent, the terms with $n<i\leq n+k-1$ are not zero. But on the other hand it looks like this is probably what OP wanted to write in the first place. $\endgroup$ – Marc van Leeuwen Jan 6 '15 at 6:27
  • $\begingroup$ @MarcvanLeeuwen Isn't the convention for positive integers $a,b$ this? $$\binom{-a}{b}=0$$ $\endgroup$ – Zubin Mukerjee Jan 6 '15 at 13:56
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    $\begingroup$ @ZubinMukerjee: No it isn't. Instead $$ \binom{-a}b=\frac{(-a)(-a-1)\ldots(-a-b+1)}{b!} = (-1)^b\binom{a+b-1}b$$ which is nonzero when $a>0$. $\endgroup$ – Marc van Leeuwen Jan 6 '15 at 14:26
  • $\begingroup$ @MarcvanLeeuwen This is mind-blowing, and makes so much sense! Does it correspond to an extension of Pascal's triangle above the first row? It reminds me of when I first realized that you could continue the Fibonacci numbers in the other direction for $$\dots, -5,3,-2,1,-1,0,1,1,2,3,5, \dots$$ $\endgroup$ – Zubin Mukerjee Jan 6 '15 at 15:42

Consider the number of ordered pairs $(a, S)$ such that $S$ is a $k$-element subset of $\{1,2, \dots, n\}$ and $a \le \min S$.

One way of counting:

Fix $\min S$.

If $\min S = i$, the number of sets = $\binom{n-i}{k-1}$ (choose $k-1$ from $\{i+1, i+2, \dots, n\}$. For each such $S$, you have $i$ possibilities for $a$.

Thus the number of $(a,S)$ pairs = $\sum_{i=1}^{n-k+1} i\binom{n-i}{k-1}$

Now count that differently: either $a = \min S$ or not.

If $a \neq \min S$, then the number is $\binom{n}{k+1}$ (pick $k+1$ elements basically)

If $a = \min S$, the number is $\binom{n}{k}$.

Thus the numerator you seek is $\binom{n}{k} + \binom{n}{k+1} = \binom{n+1}{k+1}$

So your expression simplifies to $\dfrac{n+1}{k+1}$.

(Note, I have assumed you wanted the sum upto $n-k+1$)

  • $\begingroup$ You don't seem to account for the factor $i$ in the summand. $\endgroup$ – Marc van Leeuwen Jan 6 '15 at 6:14
  • $\begingroup$ @MarcvanLeeuwen: You, are right! I missed the i. $\endgroup$ – Aryabhata Jan 6 '15 at 6:16
  • $\begingroup$ @MarcvanLeeuwen: I attempted to fix it. Thanks for pointing out the mistake. $\endgroup$ – Aryabhata Jan 6 '15 at 6:27
  • $\begingroup$ Aryabhata : Thank you very much. Can you please put the bounds on the summation symbol to make it more clear? $\endgroup$ – aghost Jan 6 '15 at 16:39

Write the numerator (after replacing the upper bound by $n$ or $n-k+1$, which I suppose was intended) as $\sum_{i=0}^n\binom i1\binom{n-i}{k-1}$. This summation then gives $\binom{n+1}{k+1}$ because of the more general formula $$ \sum_{i=0}^k\binom im\binom{k-i}n=\binom{k+1}{m+n+1} $$ that I mentioned with a combinatorial proof in this answer; put $(k,m,n):=(n,1,k-1)$ to get the special case needed here.

The proof specialises for this special case $\sum_{i=0}^n\binom i1\binom{n-i}{k-1}=\binom{n+1}{k+1}$ as follows. Locate, in the subset of $k+1$ elements out of $n+1$ for the right hand side, the second-smallest element, and let $i$ be the number of elements strictly smaller than that element. Then $\binom i1$ choices remain for the smallest element, and $\binom{n-i}{k-1}$ choices for the set of $k-1$ larger elements, and all possibilities are counted by the left hand side.

All in all your formula simplifies to $\left.\binom{n+1}{k+1}\right/\binom nk=\frac{n+1}{k+1}$.

  • $\begingroup$ See also this question and the list of questions I linked to in the comments below that question. $\endgroup$ – Marc van Leeuwen Jan 6 '15 at 14:41

I am very bad in combinatorics (as in so many other areas) so I shall not prove anything.

Intrigued by the expression (almost by the upper bound for the summation) and using a CAS, what was found is that $$\sum_{i=1}^{n+k-1} i \binom{n-i}{k-1}=\frac{(k-1) \binom{-k}{k-1} \Big(k (k+n-1)+n+1\Big)+n (n+1) \binom{n-1}{k-1}}{k (k+1)}$$ where appears a generalized binomial coefficient $\binom{-k}{k-1}$.

Hoping this could be of some use.


$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{{\ds{\sum_{j\ =\ 1}^{n\ +\ k\ -\ 1}j{n - j \choose k - 1}}\over \ds{n \choose k}}:\ {\large ?}.\qquad}$ Lets $\ds{\quad n + k - 1 \equiv m}$.

\begin{align}&\color{#66f}{\large\sum_{j\ =\ 1}^{m}j{n - j \choose k - 1}} =\sum_{j\ =\ 1}^{m}j\oint_{\verts{z}\ =\ 1^{-}}{\pars{1 + z}^{n - j} \over z^{k}} \,{\dd z \over 2\pi\ic} \\[5mm]&=\oint_{\verts{z}\ =\ 1^{-}} {\pars{1 + z}^{n} \over z^{k}}\sum_{j\ =\ 1}^{m}j\pars{1 \over z + 1}^{j} \,{\dd z \over 2\pi\ic} \\[5mm]&=\oint_{\verts{z}\ =\ 1^{-}} {\pars{1 + z}^{n} \over z^{k}}\bracks{% {1 + z \over z^{2}} - {\pars{1 + z}^{-m + 1} \over z^{2}} -m\,{\pars{1 + z}^{-m} \over z}}\,{\dd z \over 2\pi\ic} \\[5mm]&={n + 1 \choose k + 1} - {n - m + 1 \choose k + 1} - m{n - m \choose k} \\[5mm]&=\color{#66f}{\large{n + 1 \choose k + 1} - {2 - k \choose k + 1} -\pars{n + k - 1}{1 - k \choose k}}\,,\qquad \boxed{\ds{\quad n + k \geq 2\quad}} \end{align}


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