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I am to find a combinatorial argument for the following identity:

$$\sum_k \binom {2r} {2k-1}\binom{k-1}{s-1} = 2^{2r-2s+1}\binom{2r-s}{s-1}$$

For the right hand side, I was think that would just be number of ways to choose at least $s-1$ elements out of a $[2r-s]$ set. However, for the left hand side, I don't really know what it is representing.

Any help would be greatly appreciated!

Martin Sleziak
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metricspace
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4 Answers4

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Although this answer was accepted, it is in fact incorrect: the operation described in the second paragraph does not yield the same outcomes as the one described in the first paragraph. I’ve added an example below the answer.

HINT: We have $2r-s$ white balls numbered $1$ through $2r-s$. We pick $s-1$ of them and paint those balls red, and we stick gold stars on any subset of the remaining white balls; since there are $2r-s-(s-1)=2r-2s+1$ white balls remaining, there are $$\binom{2r-s}{s-1}2^{2r-2s+1}\tag{1}$$ possible outcomes of this sequence of operations.

Alternatively, we can start with $2r$ white balls numbered $1$ through $2r$. We pick a set $\mathscr{B}$ containing an odd number of these balls, $2k-1$ for some $k\in[r]$, and line them up in numerical order. The $k$-th ball in line is the one in the middle; call it $B$. We choose $s-1$ of the $k-1$ chosen balls with numbers smaller than that of $B$ and paint them red. This is possible only if $k\ge s$, in which case the number on $B$ is at most $2r-s+1$, and the red balls all have numbers in the set $[2r-s]$. We now throw away the balls in $\mathscr{B}$ that have numbers not in $[2r-s]$ and stick gold stars on any white balls left in the set of chosen balls. At this point we have $s-1$ red balls and possibly some white balls with gold stars. There are

$$\sum_k\binom{2r}{2k-1}\binom{k-1}{s-1}$$

possible outcomes.

  • Verify that these possible outcomes are exactly the same as for the first sequence of operations.

Counterexample: Let $r=3$ and $s=2$; the count in $(1)$, which is clearly correct, shows that there are $\binom41 2^3=32$ possible outcomes from the first procedure.

In the second procedure the possible values of $k$ are $2$ and $3$, and neither can produce the outcome in which ball $4$ is red, ball $1$ has a star, and balls $2$ and $3$ are plain white.

  • If $k=5$, the number on $B$ is at most $4$, so the only balls that can possibly end up red are balls $1,2$, and $3$.
  • If $k=3$ and ball $4$ is red, $B$ must be ball $5$, $\mathscr{B}$ must comprise balls $4,5$, and $6$, and no ball gets a star.

(It’s also not hard to verify that the arrangement in which balls $1$ and $2$ are red and balls $3$ and $4$ are starred is generated twice, once when $\mathscr{B}$ comprises balls $1,2,3,4$ and $5$ and once when it comprises balls $1,2,3,4$ and $6$.)

Brian M. Scott
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  • how could it be possible if you have $s\geq k$? Also, how did you get the bound on $B$? – metricspace Sep 07 '16 at 04:29
  • @metricspace: If $s\ge k$, $\binom{k-1}{s-1}=0$, so the $k$ term is $0$: this case simply doesn’t get counted. As for the number on $B$, there are $k-1$ chosen balls with larger numbers, and $k\ge s$ (or the $k$ term is $0$ and doesn’t matter), so there are at least $s-1$ balls with larger numbers than $B$. That means that $B$’s number is at most $2r-(s-1)=2r-s+1$. – Brian M. Scott Sep 07 '16 at 05:35
  • I don't think the bijection is clear as the answer makes it seem. For the second sequence of operations, in order to color $2r-s$ red, $B$ needs to be $2r-s+1$ (in which case $k = s$), but then all the chosen balls less than $B$ are colored red while everything else is discarded, leaving no room for gold stars, i.e. it can't produce an outcome that has $2r-s$ colored red and has something else as a gold star. The problem comes from the 'throwing away' process, but I don't see a clear fix for it. – Mean X Apr 04 '24 at 02:43
  • To see why there simply isn't a clear fix that can replace 'throwing away', we evaluate the number of configurations of balls that have a particular set of red ones for each process. For example, take $r=4, s=3, k=3$, and count the number of configurations that have $1,2$ as the red ones: there are $2r-s=5$ balls in total for the first process, thus $2^{(5-2)}=8$ configurations, while for the second process we have $8$ balls total and need to choose $3$ other ones among the remaining $6$ for a total of $\binom{6}{3}=20$ configurations. – Mean X Apr 04 '24 at 05:21
  • @MeanX: You’re right. For an even more precise illustration of the problem, note that in your example there are three ways for the second operation to produce the arrangement with $1$ and $2$ red and $3$ starred, while that operation cannot produce the arrangement with $1$ and $5$ red and $3$ starred, because if $1$ and $5$ are red, the originally chosen set of $5$ balls must have had numbers $1,5,6,7$, and $8$, since $B$ must have been number $6$. I can’t delete the answer, because it’s been accepted, but I can change it to point out the error, though I’ll give it a little more thought first. – Brian M. Scott Apr 04 '24 at 18:12
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Suppose we seek to verify that $$\sum_{k=1}^r {2r\choose 2k-1} {k-1\choose s-1} = 2^{2r-2s+1} {2r-s\choose s-1}$$

where presumably $s\ge 1$. The lower limit is set to $k=1$ as the first binomial coefficient is zero when $k=0.$

Introduce $${2r\choose 2k-1} = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{2r-2k+2}} \frac{1}{(1-z)^{2k}} \; dz.$$

This provides range control and vanishes when $k\gt r$ so we may extend the range to infinity, obtaining

$$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{2r+2}} \sum_{k\ge 1} {k-1\choose s-1} \frac{z^{2k}}{(1-z)^{2k}} \; dz.$$

This yields

$$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{2r+2}} \sum_{k\ge s} {k-1\choose s-1} \frac{z^{2k}}{(1-z)^{2k}} \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{2r+2}} \frac{z^{2s}}{(1-z)^{2s}} \sum_{k\ge 0} {k+s-1\choose s-1} \frac{z^{2k}}{(1-z)^{2k}} \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{2r+2}} \frac{z^{2s}}{(1-z)^{2s}} \frac{1}{(1-z^2/(1-z)^2)^s} \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{2r-2s+2}} \frac{1}{((1-z)^2-z^2)^s} \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{2r-2s+2}} \frac{1}{(1-2z)^s} \; dz.$$

This is

$$[z^{2r-2s+1}] \frac{1}{(1-2z)^s} = 2^{2r-2s+1} {2r-2s+1+s-1\choose s-1} \\ = 2^{2r-2s+1} {2r-s\choose s-1}$$

and we have the claim.

Addendum Apr 2024. For the geometric series to converge we need $|z^2/(1-z)^2| \lt 1.$ But with $\epsilon\ll 1$ we have $|z^2/(1-z)^2| \le \epsilon^2/(1-\epsilon)^2$ so we require $\epsilon^2 \lt (1-\epsilon)^2$ or $\epsilon \lt 1/2.$

Marko Riedel
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$$ \begin{align} \sum_k\binom{2r}{2k-1}\binom{k-1}{s-1} &=\sum_k\binom{2r}{2r-2k+1}\binom{k-1}{k-s}\tag1\\ &=\sum_k(-1)^{k-s}\binom{2r}{2r-2k+1}\binom{-s}{k-s}\tag2\\[3pt] &=\left[x^{2r-2s+1}\right](1+x)^{2r}\left(1-x^2\right)^{-s}\tag3\\[12pt] &=\left[x^{2r-2s+1}\right](1+x)^{2r-s}(1-x)^{-s}\tag4\\[9pt] &=\sum_k(-1)^k\binom{2r-s}{2r-2s-k+1}\binom{-s}{k}\tag5\\ &=\sum_k\binom{2r-s}{2r-2s-k+1}\binom{k+s-1}{k}\tag6\\ &=\sum_k\binom{2r-s}{k+s-1}\binom{k+s-1}{s-1}\tag7\\ &=\binom{2r-s}{s-1}\sum_k\binom{2r-2s+1}{k}\tag8\\ &=2^{2r-2s+1}\binom{2r-s}{s-1}\tag9 \end{align} $$ Explanation:
$(1)$: symmetry of Pascal's triangle
$(2)$: apply negative binomial coefficient
$(3)$: write the sum as the coefficient in a product
$(4)$: cancel factors
$(5)$: write the coefficient in a product as a sum
$(6)$: apply negative binomial coefficient
$(7)$: symmetry of Pascal's triangle
$(8)$: $\binom{n}{m}\binom{m}{k}=\binom{n}{k}\binom{n-k}{m-k}$
$(9)$: evaluate the sum

robjohn
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  • +1 for a nice and clean proof! :) –  Sep 13 '19 at 15:16
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    @Brahadeesh: I am working on turning steps $(3)$ and $(4)$ into a combinatorial argument. If it is significantly different from Brian Scott's, I will amend my answer. – robjohn Sep 13 '19 at 15:33
  • Do ping me if you succeed, I'd love to see such an argument :) –  Sep 13 '19 at 15:38
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Since there isn't a correct combinatorial interpretation currently, I have done robjohn's unfinished work of turning their answer into one.
The idea is to use the following:

$$[x^k](1+x)^n = \text{# of ways to choose } k \text{ elements from a set of }n = \binom{n}{k}$$ $$[x^k](1-x)^{-n} = [x^k](1+x+x^2+\cdots)^n = \text{# of ways to express } k \text{ as a sum of } n \text{ nonnegative integers} = \binom{n+k-1}{k} $$

I reiterate here the main steps that will be translated:

\begin{align*} \sum_k\binom{2r}{2k-1}\binom{k-1}{s-1} &=\sum_k\binom{2r}{2r-2k+1}\binom{k-1}{k-s}\\ &=\left[x^{2r-2s+1}\right](1+x)^{2r}\left(1-x^2\right)^{-s}\tag1\\[12pt] &=\left[x^{2r-2s+1}\right](1+x)^{2r-s}(1-x)^{-s}\tag2\\[9pt] &=\sum_l\binom{2r-s}{2r-2s-l+1}\binom{l+s-1}{l}\tag3\\ &=\sum_l\binom{2r-s}{s-1}\binom{2r-2s+1}{l}\tag4\\ &=2^{2r-2s+1}\binom{2r-s}{s-1}\\ \end{align*}

We give steps $(1)$ and $(4)$ the following interpretations, respectively:

$A$. Given $2r$ numbered balls with indices $i = 1, \cdots, 2r$, assign labels $a_i=0$ to $2k-1$ of them and $a_i = 1$ to the rest. To balls with index $j > 2r-s$, assign a second nonnegative label $b_j$ such that $\displaystyle\sum_j b_j = k-s$.

Here, the $a_i$'s correspond to the term chosen in the $i$-th $(x+1)$ in the expansion of $(x+1)^{2r}$. Likewise, the $b_j$'s correspond to half of the $(j-(2r-s))$-th integer in a list of $s$ nonnegative even integers that sum to $2k-2s$.

$B$. Given $2r-s$ numbered balls, assign starred labels $0\star$ to $s-1$ of them. Among the remaining unlabeled ones, assign labels $0$ to $l$ of them and $1$ to the rest.

This is equivalent to the first sequence of operations given in Brain M. Scott's answer.

The bijection between the two processes are given by steps $(2)$ and $(3)$.

To go from $A$ to $B$, observe that performing the multiplication $(1+x)^{s}(1-x^2)^{-s}$ is equivalent to merging the two indices $a_j+2b_j$ for each $j>2r-s$: call this merged index $c_j$. Note:
$$2r-2s+1 = \sum_{i} a_i + \sum_{j} 2b_j = \sum_{i\leq2r-s} a_i + \sum_{j} c_j$$ Let $l = \displaystyle\sum_{j} c_j$ such that $\displaystyle\sum_{i\leq2r-s} a_i = 2r-2s-l+1$, i.e. there are exactly $s+l-1$ $0$'s among the remaining (unmerged) $a_i$'s. To go from $(3)$ to $B$, we assign stars$(\star)$ to $s-1$ of the $0$'s such that each $c_j$ is the number of $0$'s between the $(j-(2r-s)-1)$-th and $(j-(2r-s))$-th starred balls, where the first $0\star$ is preceded by $c_{2r-s+1}$ $0$'s and the last $0\star$ is succeeded by $c_{2r}$ $0$'s. The set of unmerged balls gives us the description of $B$.

Now, to go from $B$ into $A$ we simply revert the above process, obtaining the merged labels $c_j$ from the configuration of $0$'s and unmerging them, noting that $a_j$ is determined by the parity of $c_j$.

As an illustration, take $r=9, s=4$.

This is a configuration of $A$ with $k=7$, thus there are $2k-1 = 13$ balls that are assigned $a_i = 0$, and the $b_j$'s sum to $k-s = 3$, denoted as subscripts of $a_j$. The balls have been separated after index $2r-s = 14$ for clarity.

$$1\ 0\ 0\ 0\ 0\ 1\ 0\ 0\ 0\ 1\ 0\ 0\ 0\ 0\ |\ 1_2\ 0_0\ 1_0\ 0_1$$

Merging $c_j = a_j + 2b_j$ yields:

$$1\ 0\ 0\ 0\ 0\ 1\ 0\ 0\ 0\ 1\ 0\ 0\ 0\ 0\ |\ 5\ \ 0\ \ 1\ \ 2$$

Finally, starring the unmerged $0$'s on the left according to the $c_j$'s, we have our configuration of $B$ with $s-1 = 3$ $0\star$'s and $l = \displaystyle\sum_{j} c_j = 8$ unstarred $0$'s.

$$1\ 0\ 0\ 0\ 0\ 1\ 0\ 0\!\star 0\!\star 1\ 0\ 0\!\star 0\ 0$$

Mean X
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