Proving in combinations ... is it true ?

Question

prove that :

$$\sum_{k=0}^n \left ( \begin{array}{c}n\\k\end{array}\right )^2 = {2n \choose n}$$

I tried to prove it by supposing i have a class has (M) girls and n boys , and we need to choose a team (K of students) ---- So the ways are $$\displaystyle {m+n \choose k}$$---------- So I have (0) boy and (k) girl OR 1 of n and (k-1) of m OR 2 of n and (k-2) of m OR ... (k) of n and (0) of m

SO:

$(^{m}_{0})(^n_k)+(^m_1)(^{n}_{k-1})+(^{m}_{2})(^{n}_{k-2})+...(^{m}_{k})(^{n}_{0})$

THEN

$$ \ \sum_{i=0}^k{m \choose i}{n \choose k-i}\ $$

$$ \ \sum_{i=0}^k{m \choose i}{n \choose k-i} ={n+m \choose k}\ $$

suppose m=n=k (is it true to suppose this ? )

$$ \ \sum_{i=0}^n{n \choose i}{n \choose n-i} ={2n \choose n}\ $$

i+(n-i)=2n the below = the top

what should i do ??

Can i prove it by Mathematical induction ?

This is incredibly confusing. You should clean it up a bit. Tell us what parts you have done, what equation you finished off on, and what problem you are having exactly. — Cameron Williams, Apr 13 '17 at 17:58
There are much elegant and beautiful ways to prove this, i think induction is just last approach if you end up no where anyways you can prove it by induction just try and use binomial identities on n — Ash Pd, Apr 13 '17 at 18:03
If any one can prove it by mathematical induction and send it as a picture — Hamzeh K. Yacoub, Apr 13 '17 at 18:05
I strongly discourage you from attempting to do this via mathematical induction, it will not be pretty. If you insist though, see this post. — JMoravitz, Apr 13 '17 at 18:11

score 1 · Answer 1 · answered Apr 13 '17 at 18:02

Imagine we want to solve this question. "We have $2n$ objects that we want to chose $n$ of them. How many states exist?"

The first thing we can think of is this: $\binom{2n}{n}$

But we can solve it in another way. We can divide these objects into two parts each consisted of $n$ objects. Now we pick $k$ objects from the first group and the rest $n - k$ from the second group. So we will have $\binom{n}{k} * \binom{n}{n-k}$ possibilities. This equation is the equivalent to:

$\binom{n}{k} * \binom{n}{n - k} = \binom{n}{k} * \binom{n}{k} = \binom{n}{k}^2$

But we have to calculate these possibilities for all $k$. It is obvious that $k$ is between $0$ and $n$. So the answer will be:

$\sum_{k=0}^{n}{\binom{n}{k}}^2$

So we have two solutions for the same problem. Since this question has only one answer, these solutions must be equal. So:

$\sum_{k=0}^{n}{\binom{n}{k}}^2 = \binom{2n}{n}$

score 1 · Answer 2 · answered Apr 13 '17 at 18:03

1

Another approach :

The coefficient of $x^n$ in the product : $$(1+x)^n(x+1)^n$$

Is equal to the sum : $\displaystyle \left ( \begin{array}{c}n\\0\end{array}\right )\left ( \begin{array}{c}n\\0\end{array}\right )+\left ( \begin{array}{c}n\\1\end{array}\right )\left ( \begin{array}{c}n\\1\end{array}\right )+ \dots=\sum_{k=0}^n \left ( \begin{array}{c}n\\k\end{array}\right )^2 $

Since this product is equal to : $(1+x)^{2n}$, coefficient of $x^n$ must be same.

From here we get coefficient of $x^n$ as $\displaystyle{2n \choose n}$.

Hence : $$\sum_{k=0}^n \left ( \begin{array}{c}n\\k\end{array}\right )^2 = {2n \choose n}$$

answered Apr 13 '17 at 18:03

Jaideep Khare

19,293

thank u , i didn't understand all of it – Hamzeh K. Yacoub Apr 13 '17 at 18:19
What part?Can you elaborate – Jaideep Khare Apr 13 '17 at 18:24
What is the coefficient in the product ? And how is it equal to the sum of (....) ? – Hamzeh K. Yacoub Apr 13 '17 at 18:27
@HamzehK.Yacoub See this :https://math.stackexchange.com/a/2195308/421580 thread as this is closely related to my this answer. I am pretty sure you'll understand that completely. – Jaideep Khare Apr 13 '17 at 18:27
By coefficient I mean coefficient is binomial expansion. – Jaideep Khare Apr 13 '17 at 18:28

Proving in combinations ... is it true ?

$$\sum_{k=0}^n \left ( \begin{array}{c}n\\k\end{array}\right )^2 = {2n \choose n}$$

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