Combinatorial proofs of identities use double counting and combinatorial characterizations of binomial coefficients, powers, factorials etc. They avoid complicated algebraic manipulations.
Questions tagged [combinatorial-proofs]
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Identity for convolution of central binomial coefficients: $\sum\limits_{k=0}^n \binom{2k}{k}\binom{2(n-k)}{n-k}=2^{2n}$
It's not difficult to show that
$$(1-z^2)^{-1/2}=\sum_{n=0}^\infty \binom{2n}{n}2^{-2n}z^{2n}$$
On the other hand, we have $(1-z^2)^{-1}=\sum z^{2n}$. Squaring the first power series and comparing terms gives us
$$\sum_{k=0}^n…
Skatche
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Combinatorial proof of summation of $\sum\limits_{k = 0}^n {n \choose k}^2= {2n \choose n}$
I was hoping to find a more "mathematical" proof, instead of proving logically $\displaystyle \sum_{k = 0}^n {n \choose k}^2= {2n \choose n}$.
I already know the logical Proof:
$${n \choose k}^2 = {n \choose k}{ n \choose n-k}$$
Hence summation can…
Lance C
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Combinatorial proof that $\sum \limits_{k=0}^n \binom{2k}{k} \binom{2n-2k}{n-k} (-1)^k = 2^n \binom{n}{n/2}$ when $n$ is even
In my answer here I prove, using generating functions, a statement equivalent to
$$\sum_{k=0}^n \binom{2k}{k} \binom{2n-2k}{n-k} (-1)^k = 2^n \binom{n}{n/2}$$
when $n$ is even. (Clearly the sum is $0$ when $n$ is odd.) The nice expression on the…
Mike Spivey
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The Hexagonal Property of Pascal's Triangle
Any hexagon in Pascal's triangle, whose vertices are 6 binomial coefficients surrounding any entry, has the property that:
the product of non-adjacent vertices is constant.
the greatest common divisor of non-adjacent vertices is constant.
Below is…
milcak
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Two interview questions
I recently came across two interview questions for admission in B.Math at an university. I gave the two questions a try and want to know if my solutions are correct or not.
Q1: Given that $x^4-4x^3+ax^2+bx+1=0$ has all positive roots and $a,b\in\Bbb…
idiot
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Is there a combinatorial interpretation of the triangular numbers?
The triangular numbers count the number of items in a triangle with $n$ items on a side, like this:
This can be calculated exactly by the formula $T_n = \sum_{k=1}^n k = \frac{n(n+1)}{2} = {n+1 \choose 2} = {n+1 \choose n-1}$.
Is there any…
sagittarian
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Combinatorial interpretation of sum of squares, cubes
Consider the sum of the first $n$ integers:
$$\sum_{i=1}^n\,i=\frac{n(n+1)}{2}=\binom{n+1}{2}$$
This makes the following bit of combinatorial sense. Imagine the set $\{*,1,2,\ldots,n\}$. We can choose two from this set, order them in decreasing…
2'5 9'2
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Evaluate $\sum\limits_{k=1}^n k^2$ and $\sum\limits_{k=1}^n k(k+1)$ combinatorially
$$\text{Evaluate } \sum_{k=1}^n k^2 \text{ and } \sum_{k=1}^{n}k(k+1) \text{ combinatorially.}$$
For the first one, I was able to express $k^2$ in terms of the binomial coefficients by considering a set $X$ of cardinality $2k$ and partitioning it…
kuch nahi
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Show by combinatorial argument that ${2n\choose 2} = 2{n \choose 2} + n^2$
So i was given this question. Show by combinatorial argument that ${2n\choose 2} = 2{n \choose 2} + n^2$
Here is my solution:
Given $2n$ objects, split them into $2$ groups of $n$, $A$ and $B$. $2$-combinations can either be assembled both from…
Zero
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Combinatorial proof of $\sum\limits_{k=0}^n {n \choose k}3^k=4^n$
Using the following equation:
$$\sum_{k=0}^n {n \choose k}3^k=4^n$$
I need to prove that both sides of the equation solve the same combinatorial problem.
It's easy to see that the right side of the equation is counting number of ways to divide $n$…
MichaelS
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Strange inequality involving *nested* binomial coefficients and combinatorial interpretation
Recently, I solved a question that asked for a geometric proof of the identity $\binom{\binom{n}{3}}{2} < \binom{\binom{n}{2}}{3}$ and I did it here using the combinatorial interpretation of lines and triangles in an $n$-gon, and exploiting the…
Sarvesh Ravichandran Iyer
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Is there a combinatorial proof that $e$ is finite?
I'm looking for an integer $N$ and a combinatorial proof either that $(n+1)^n
MTyson
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How to show $\sum_{k=0}^{n}\binom{n+k}{k}\frac{1}{2^k}=2^{n}$
How does one show that $$\sum_{k=0}^{n}\binom{n+k}{k}\frac{1}{2^k}=2^{n}$$ for each nonnegative integer $n$?
I tried using the Snake oil technique but I guess I am applying it incorrectly. With the snake oil technique we have $$F(x)=…
C.S.
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Proving $k \binom{n}{k} = n \binom{n-1}{k-1}$
Suppose we want to prove $$ k \binom{n}{k} = n \binom{n-1}{k-1}$$
In the LHS we are choosing a team of $k$ players from $n$ players. Then we are choosing a captain. In the RHS we are choosing a captain from the $n$ players. Then we are choosing the…
James
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Show that $\sum\limits_{k=0}^n\binom{2n}{2k}^{\!2}-\sum\limits_{k=0}^{n-1}\binom{2n}{2k+1}^{\!2}=(-1)^n\binom{2n}{n}$
How can I prove the identity:
$$
\sum_{k=0}^n\binom{2n}{2k}^2-\sum_{k=0}^{n-1}\binom{2n}{2k+1}^2=(-1)^n\binom{2n}{n}?
$$
Maybe, can we expand
$$
f(x)=(1+x)^{2n}?
$$
Thank you.
math110
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