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Questions tagged [combinatorial-proofs]

Combinatorial proofs of identities use double counting and combinatorial characterizations of binomial coefficients, powers, factorials etc. They avoid complicated algebraic manipulations.

2 votes
0 answers
62 views

This is a proof for the hermite identity: $$\sum_{k=0}^{n-1} \left\lfloor x + \frac{k}{n} \right\rfloor=\lfloor nx \rfloor$$ Combinatorial Interpretation of the RHS Let $S$ be the set of positive ...
Ciderate's user avatar
2 votes
1 answer
117 views

Give a combinatorial proof that \begin{eqnarray*} \sum_{i+j+k=n} \binom{2i}{i} \binom{2j}{j} \binom{2k}{k} = (2n+1) \binom{2n}{n}. \end{eqnarray*} Where did this come from ? ... In this question (...
Donald Splutterwit's user avatar
3 votes
0 answers
111 views

Let $(a_n)_{n\ge 0}$ be defined by the tribonacci recurrence $$ a_n = a_{n-1}+a_{n-2}+a_{n-3}\qquad (n\ge 3). $$ I am looking for purely combinatorial proofs (no induction) of the following bounds: ...
Luis Eduardo Solano Noreña's user avatar
11 votes
0 answers
219 views

While playing with the Shapiro's convolution identity, I came up with the following identity: $$ \sum_{k=0}^{n} \frac{(2k)!(2n-2k)!}{k!(n-k)!n!} = \sum_{k=0}^{n} \binom{2k}{k}2^{n-k}. \tag{*} $$ This ...
Sangchul Lee's user avatar
2 votes
0 answers
89 views

Consider two sets $A$ and $B$. Let $A$ be the set of all strings of length $2n$ with exactly $n$ $(-1)$'s and $n$ $(1)$'s. Let $B$ be the set of all strings of length $2n$ having each character to be ...
Rishit Garg's user avatar
-1 votes
1 answer
62 views

Let $P(n,k)=\frac{n!}{(n-k)!}$ denote the $k$-permutation of $n$. Suggest a combinatorial proof of $$P(n+1,k)=k\sum_{i=k-1}^nP(i,k-1).$$ This identity relates the $k$-permutation of $n+1$ to the ...
James Yuen's user avatar
18 votes
1 answer
381 views

I'm looking for a combinatorial or probabilistic proof for the following beautiful identity involving the Catalan numbers $C_n$: $$ \sum_{k=0}^n C_{2k}C_{2n-2k}=2^{2n}C_n, \tag{1} $$ This has a simple ...
Sangchul Lee's user avatar
1 vote
0 answers
141 views

Context: Mircea Dan Rus's 2025 paper Yet another note on notation (a spiritual sequel to Knuth's 1991 paper Two notes on notation) introduces the syntax $x^{\{n\}}=x!{n\brace x}$ to denote the number ...
DroneBetter's user avatar
0 votes
0 answers
62 views

I was studying a general way to solve problems like choose 4 words from word 'parallel' , or more generally choose r words from n words (where it may be distinct or identical ) and I came across r ...
amgine's user avatar
  • 1
0 votes
0 answers
126 views

Let $A$ and $B$ be two totally ordered finite sets having, respectively, $|A|=p$ and $|B|=q$ elements with $p > q$. The set $B^{A}$ of all functions $f\colon A \to B$ has $|B^{A}|=q^{p}$ elements, ...
Jotazuma's user avatar
  • 162
5 votes
2 answers
201 views

Given an unlabeled plane rooted tree whose vertices only have $0,1$ or $3$ child nodes, and the number of the vertices that have $0$ or $1$ child node is $n$. Find the number $H(n)$ of trees with ...
BLUEJOKE's user avatar
3 votes
1 answer
175 views

I'm trying to figure out the value of the following sum, $S_n$. It's defined over a set $H_n$ which contains integer partitions of $n$ using only parts of size 2 or greater. Here are the definitions: $...
匚ㄖㄥᗪ乇ᗪ's user avatar
0 votes
0 answers
104 views

prove the following binomial identities $$ \sum_{k=0}^{n-r} \binom{r+k}{r} = \binom{n+1}{r+1}$$ $$ \sum_{k=0}^{m} \binom{m}{m-k} \binom{n}{r+k} = \binom{m+n}{m+r} $$ With n ≥ r +m $$ \sum_{k = n-s}^{r-...
Haekal Dinova's user avatar
1 vote
0 answers
95 views

In the proof of Theorem 4.1 of the book Random Graphs by B. Bollobas. Where it is written that Suppose the graph $A$ and $B$ are such that $B$ is an $H$-graph and it has exactly $t$ vertices not ...
Cantor_Set's user avatar
  • 1,256
2 votes
0 answers
83 views

The binomial formula $(x+y)^{n}=\sum_{k=0}^{n}\binom{n}{k}x^{k}y^{n-k}$ could be interpreted, when $x,y$ are positive integers, as a two-way counting of "words" of length $n$ which use ...
digital-Ink's user avatar
  • 2,105

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