Combinatorial proofs of the following identities

Question

I've been trying to find combinatorial proofs of the following two identities:

1: $\displaystyle\sum_{i=0}^{k} \binom{n}{i} = \sum_{i=0}^{k} \binom{n-1-i}{k-i} 2^i$ with $0 \le k \le n-1$

2: $\displaystyle\binom{2m}{2n} = \sum_{k=0}^{n} \binom{2n+1}{2k+1} \binom{m+k}{2n}$

---

For 1: The LHS is counting the number of subsets of size at most k from a set of size n. The $2^i$ in the RHS makes me think of partitioning based on what elements can be considered from the full set then either including them or not, but I can't think of a way of doing this partition without overcounting and trying to interpret the binomial term hasn't helped.

For 2: Again, the LHS is simple enough but I'm lost on how to interpret the RHS. Just from looking at it I feel like I should be considering some parity argument but haven't come up with anything else.

---

Any suggestions on how to proceed? Should I be looking for a more formal bijection?

Answer 1

To get you started, here’s a HINT for the first identity:

• Show that $\binom{n-1-i}{k-i}2^i$ is the number of subsets $A$ of $[n]=\{1,\ldots,n\}$ of cardinality at most $k$ such that $i+1\notin A$, and $|A\setminus[i]|=k-i$. That is $i+1$ is not in $A$, and $A$ has $k-i$ elements bigger than $i$.

• Suppose that $A\subseteq[n]$ has at most $k$ elements, and let $d=k-|A|$. Let $i$ be the largest integer such that $|[i]\setminus A|=d$. If $d=0$, for example, this means that $i+1$ is the smallest member of $[n]$ not in $A$. If $d=1$, $i+1$ is the second-smallest member of $[n]$ not in $A$. Show that such $i$ always exists. (Clearly it’s uniquely determined by $A$ if it does exist.)

I’ll have to think further about the second one.

Answer 2

Suppose we seek to show that

$${2m\choose 2n} = \sum_{k=0}^n {2n+1\choose 2k+1} {m+k\choose 2n}.$$

where $m\ge n.$ We introduce

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

Observe that this vanishes when $k\gt n$ so that we may use it to control the range and extend $k$ to infinity. We also use

$${m+k\choose 2n} = \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{1}{w^{2n+1}} (1+w)^{m+k} \; dw.$$

We thus obtain

$$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{2n+1}}{z^{2n+1}} \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{(1+w)^{m}}{w^{2n+1}} \sum_{k\ge 0} z^{2k} (1+w)^k \; dw\; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{2n+1} }{z^{2n+1}} \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{(1+w)^{m}}{w^{2n+1}} \frac{1}{1-(1+w)z^2} \; dw\; dz.$$

Evalute the inner integral using the negative of the residue at the pole at $$w=\frac{1-z^2}{z^2}$$ (residues sum to zero) as in

$$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{2n+1} }{z^{2n+1}} \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{(1+w)^{m}}{w^{2n+1}} \frac{1}{1-z^2 - wz^2} \; dw\; dz \\ = - \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{2n+1} }{z^{2n+3}} \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{(1+w)^{m}}{w^{2n+1}} \frac{1}{w-(1-z^2)/z^2} \; dw\; dz.$$

The negative of the residue is

$$\frac{1}{z^{2m}} \frac{z^{4n+2}}{(1-z^2)^{2n+1}} = \frac{1}{z^{2m-4n-2}} \frac{1}{(1-z^2)^{2n+1}}$$

and we obtain from the outer integral

$$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{2n+1} }{z^{2n+3}} \frac{1}{z^{2m-4n-2}} \frac{1}{(1-z^2)^{2n+1}} \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{2m-2n+1}} \frac{1}{(1-z)^{2n+1}} \; dz \\ = {2m-2n+2n\choose 2n} = {2m\choose 2n}.$$

This is the claim.

Remark. We also need to show that the contribution from the residue at infinity of the inner integral is zero. We get

$$\mathrm{Res}_{w=\infty} \frac{(1+w)^{m}}{w^{2n+1}} \frac{1}{1-(1+w)z^2} \\ = - \mathrm{Res}_{w=0} \frac{1}{w^2} (1+1/w)^{m} w^{2n+1} \frac{1}{1-z^2-z^2/w} \\ = - \mathrm{Res}_{w=0} (1+w)^{m} w^{2n-m} \frac{1}{w(1-z^2)-z^2}.$$

No contribution when $2n\ge m.$ Otherwise,

$$\frac{1}{z^2} \mathrm{Res}_{w=0} (1+w)^{m} \frac{1}{w^{m-2n}} \frac{1}{1-w(1-z^2)/z^2} \\ = \frac{1}{z^2} \sum_{q=0}^{m-2n-1} {m\choose m-2n-1-q} \frac{(1-z^2)^q}{z^{2q}} \\ = \frac{1}{z^2} \sum_{q=0}^{m-2n-1} {m\choose 2n+1+q} \left(\frac{1}{z^2}-1\right)^q$$

Combining this with the integral in $z$ yields

$$\sum_{q=0}^{m-2n-1} {m\choose 2n+1+q} \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{2n+1} }{z^{2n+1}} \frac{1}{z^2} \sum_{p=0}^q {q\choose p} (-1)^{q-p} \frac{1}{z^{2p}} \; dz.$$

The contribution from the residue is

$$[z^{2n+2+2p}] (1+z)^{2n+1} = 0.$$

We can express this verbally by saying that the term from the integral is $[z^{2n}] (1+z)^{2n+1} = 0 $ and the sum only contributes negative powers of $z$ with exponent starting at two.

Remark, II. From the convergence we require that $|z^2(1+w)| < 1$ in the double integral and must choose our contours appropriately. We must also verify that $(1-z^2)/z^2$ is outside the contour $|w|=\gamma.$ This is $1/z^2-1$ i.e. a circle of radius $1/\epsilon^2$ shifted by one to the left. Therefore when $\epsilon < 1/\sqrt{2}$ the pole is outside the contour.

Answer 3

Since a proof of my identity has been given, I shall give a proof which is partly combinatorial. The combinatorial part is as follows. For integers $a,b$ such that $a\leq b$, write $$[a,b]:=\{a,a+1,\ldots,b-1,b\}\,.$$ Consider a subset $S\subseteq [1,2m]$ of size $2n$. There are $\dbinom{2m}{2n}$ ways to choose such subsets. Write $S_1:=S\cap[1,m]$ and $$S_2:=\big(S\cap[m+1,2m]\big)-m=\big\{s-m\,\big|\,s\in S\cap[m+1,2m]\big\}\,.$$ We are counting the number of subsets $S$ of $[1,2m]$ of size $2n$ with $\left|S_1\cap S_2\right|=j$ for each $j=0,1,2,\ldots,n$.

First, there are $\dbinom{m}{2n-j}$ ways to choose $S_1\cup S_2$. Amongst the $2n-j$ chosen numbers, we can choose $S_1\cap S_2$ in $\dbinom{2n-j}{j}$ ways. That leaves $2(n-j)$ elements each of which can either belong only in $S_1$ or only in $S_2$. Thus, $$\binom{2m}{2n}=\sum_{j=0}^n\,\binom{m}{2n-j}\,\binom{2n-j}{j}\,2^{2(n-j)}\,.\tag{1}$$

We shall now prove that, for integers $M,N,K$ with $0\leq K\leq N\leq M$, we have $$\binom{M}{N-K}=\sum_{i=0}^K\,(-1)^i\,\binom{K}{i}\,\binom{M+K-i}{N}\,.\tag{2}$$ The left-hand side is the number of ways to choose $N$ elements from $[1,M+K]$ such that every number in $[M+1,M+K]$ is selected. The right-hand side is a direct result of the Principle of Inclusion and Exclusion, noting that $\dbinom{K}{i}$ is the number of ways to select $i$-subsets $T$ of $[M+1,M+K]$ and $\dbinom{M+K-i}{N}$ is precisely the number of ways to choose an $N$-subset of $[1,M+K]\setminus T$.

From (1) and (2), we get $$\binom{2m}{2n}=\sum_{j=0}^n\,\sum_{i=0}^j\,(-1)^{i}\,\binom{j}{i}\,\binom{m+j-i}{2n}\,\binom{2n-j}{j}\,2^{2(n-j)}\,.$$ Let $k:=j-i$ and, by reindexing, we have $$\binom{2m}{2n}=\sum_{k=0}^n\,\binom{m+k}{2n}\,\sum_{j=k}^n\,(-1)^{j-k}\,\binom{j}{k}\,\binom{2n-j}{j}\,2^{2(n-j)}\,.$$ This is where my identity (now with a combinatorial proof---at least partially) comes in, and we are done with $$\binom{2m}{2n}=\sum_{k=0}^n\,\binom{m+k}{2n}\,\binom{2n+1}{2k+1}\,.$$

Answer 4

Here is another variation based upon the usage of the coefficient of operator $[z^k]$ to denote the coefficient of $z^k$ in a series. This way we can write e.g. \begin{align*} [z^k](1+z)^n=\binom{n}{k} \end{align*}

We start with (2) and obtain for $0\leq n\leq m$ \begin{align*} \sum_{k=2n-m}^m&\binom{2n+1}{2k+1}\binom{m+k}{2n}\tag{1}\\ &=\sum_{k=0}^{2m-2n}\binom{2n+1}{4n-2m+2k+1}\binom{2n+k}{2n}\tag{2}\\ &=\sum_{k=0}^{2m-2n}\binom{2n+1}{2m-2n-2k}\binom{-(2n+1)}{k}(-1)^k\tag{3}\\ &=\sum_{k=0}^{\infty}[z^{2m-2n-2k}](1+z)^{2n+1}[u^{k}](1+u)^{-(2n+1)}(-1)^k\tag{4}\\ &=[z^{2m-2n}](1+z)^{2n+1}\sum_{k=0}^\infty\left(-z^2\right)^k[u^k](1+u)^{-(2n+1)}\tag{5}\\ &=[z^{2m-2n}](1+z)^{2n+1}(1-z^2)^{-(2n+1)}\tag{6}\\ &=[z^{2m-2n}](1-z)^{-(2n+1)}\\ &=[z^{2m-2n}]\sum_{k=0}^\infty\binom{-(2n+1)}{k}(-z)^k\tag{7}\\ &=\binom{-(2n+1)}{2m-2n}\tag{8}\\ &=\binom{2m}{2n}\tag{9} \end{align*} and the claim follows.

Comment:

• In (1) we sum starting from $k=2n-m$, since $\binom{m+k}{2n}=0$ for $0\leq k<2n-m$.

• In (2) we shift the index $k$ to start from zero.

• In (3) we use the binomial identities $\binom{p}{q}=\binom{p}{p-q}$ and $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.

• In (4) we apply the coefficient of operator twice and extend to upper limit of the series to $\infty$ without changing anything since we are adding zeros only.

• In (5) we use the linearity of the coefficient of operator and apply the rule $[z^{p+q}]A(z)=[z^p]z^{-q}A(z)$.

• In (6) we apply the substitution rule with $u=-z^2$ \begin{align*} A(z)=\sum_{k=0}^\infty a_kz^k=\sum_{k=0}^\infty z^k [u^k]A(u) \end{align*}

• In (7) we use the binomial series expansion

• In (8) we select the coefficient of $z^{2m-2n}$.

• In (9) we apply the identity $\binom{-p}{q}=\binom{p+q-1}{p-1}(-1)^q$.

Using the same technique we show (1): \begin{align*} \sum_{i=0}^k\binom{n}{i}=\sum_{i=0}^k\binom{n-1-i}{k-i}2^i \end{align*}

We obtain \begin{align*} \sum_{i=0}^k\binom{n-1-i}{k-i}2^i &=\sum_{i=0}^\infty[z^{k-i}](1+z)^{n-1-i}2^i\\ &=[z^k](1+z)^{n-1}\sum_{i=0}^\infty\left(\frac{2z}{1+z}\right)^i\\ &=[z^k](1+z)^{n-1}\frac{1}{1-\frac{2z}{1+z}}\\ &=[z^k](1+z)^{n}\frac{1}{1-z}\\ &=[z^k]\sum_{i=0}^\infty z^i(1+z)^{n}\\ &=\sum_{i=0}^k [z^{k-i}](1+z)^{n}\\ &=\sum_{i=0}^k \binom{n}{k-i}\\ &=\sum_{i=0}^k \binom{n}{i}\\ \end{align*}

and the claim follows.