Linked Questions

7
votes
2answers
14k views

Sum of square binomial coefficients [duplicate]

Please feel free to close this is necessary as I didn't see exactly this question (some variations that I tried but didn't seem to apply. Prove: $$\sum_{k=0}^{n}{\binom{n}{k}^2}=\binom{2n}{n}$$ I ...
1
vote
3answers
554 views

Prove that:$\binom {2n}{n}$=$\sum_{r=0}^n [\binom nr ]^2$ [duplicate]

Prove that : $$\binom {2n}{n}=\sum_{r=0}^n \left[\binom nr\right]^2.$$ First of all, I tried to do in the principle of mathematical induction but I failed. Next, I expressed the binomial in algebraic ...
1
vote
3answers
105 views

Is there an expression for the sum of $\binom nr^2$ for each $n$? [duplicate]

Is there a standard expression for $$\sum_{r=0}^{n}\binom nr^2$$
2
votes
2answers
194 views

Simplify the expression - sum of squares of binomial coefficients [duplicate]

I need to simplify this thing: $$\sum_{k=0}^n{n\choose k}^2$$ I spent more than an hour thinking about it and the set of working solutions that sprang to my mind was unfortunately still empty. Then, ...
0
votes
4answers
330 views

How to prove that the sum of squared binomials equals $\binom{2n}{n}$ [duplicate]

I've stumbled upon this lemma a few times in my textbook: $$\sum_{k=0}^{n}\begin{pmatrix}n\\k\end{pmatrix}^2=\begin{pmatrix}2n\\n\end{pmatrix}$$ I've been trying to prove it, but I simply can't seem ...
1
vote
2answers
153 views

How to prove that $\sum_{i=0}^n\binom{n}{i}^2 =\binom{2n}{n}$ [duplicate]

How can I prove that $\sum_{i=0}^n\binom{n}{i}^2 =\binom{2n}{n}$ Thanks in advance!
-1
votes
2answers
68 views

What is the proof that the two below are equal? [duplicate]

Can you help me to prove the following equality ?
63
votes
6answers
50k views

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 = {...
5
votes
6answers
12k views

Proof of a binomial identity $\sum_{k=0}^n {n \choose k}^{\!2} = {2n \choose n}.$

Prove that $$\sum_{k=0}^n {n \choose k}^{\!2} = {2n \choose n}.$$ The exercise provides the following hint: $\,\,\displaystyle{n \choose k}={n\choose n-k}$. Any help?
12
votes
3answers
2k views

How to get ${n \choose 0}^2+{n \choose 1}^2+{n \choose 2}^2+\cdots+{n \choose n}^2 = {x \choose y}$

I found this in my test book, any hints? Given $${n \choose 0}^2+{n \choose 1}^2+{n \choose 2}^2+\cdots+{n \choose n}^2 = {x \choose y}$$ Then find the value of x and y in n. According to the answer ...
4
votes
4answers
4k views

Proving $ \binom n 0 ^2 + \binom n 1 ^2 + \dots + \binom n n ^2 = \binom { 2n} n $ without induction [duplicate]

I have to prove that: $$ \binom n 0 ^2 + \binom n 1 ^2 + \dots + \binom n n ^2 = \binom { 2n} n $$ I don't want a complete solution, but only a hint.
3
votes
3answers
3k views

N women and N men. Groups of pairs.

So we have $N$ men and $N$ women. We are creating groups of pairs. It is not necessary to use every man and woman. How many groups can we make ? So if we number them from $1$ to $N$ - let $W_{1}$ be ...
3
votes
6answers
112 views

Prove that $2^n(n!)^2 \leq (2n)!$

Prove that $2^n(n!)^2 \leq (2n)!$ One can also use the following result to prove the above: $2 · 6 · 10 · 14 · · · · · (4n − 2) = \frac{(2n)!}{ n!}$. The above relation gives, $(2n)!=2^n n! (1.3.5.....
1
vote
2answers
427 views

Computing $C_0^2+C_1^2+C_2^2+C_3^2+ \cdots +C_n^2$ [duplicate]

If $C_k$ denotes binomial coefficient of choosing $k$ objects from a set of $n$ objects how to calculate this: $$C_0^2+C_1^2+C_2^2+C_3^2+\cdots +C_n^2$$
3
votes
2answers
76 views

Calculate $\sum_{i=0}^{n}i^2 \cdot \binom{n}{i}^2 $

Calculate $$\sum_{i=0}^{n}i^2 \cdot \binom{n}{i}^2 $$ I tried to do this via combinatorics interpretation. So we want $$ \langle A,a,B,b \rangle $$ such that $$ A \subset \left\{ 1,...,n \right\} \\ ...

15 30 50 per page