Binomial theorem proof question

Question

enter image description hereUse the binomial theorem to prove that $(nC0)^2 + (nC1)^2 + (nC2)^2 + ... +(nCn)^2 = 2nCn$ left hand side shall be $ (2^n)^2 $ because $(nC0) + (nC1) + (nC2) + ... +(nCn) = 2^n $ and $(nC0)^2 + (nC1)^2 + (nC2)^2 + ... +(nCn)^2 = [(nC0) + (nC1) + (nC2) + ... +(nCn)]^2$ but if u tried to substitute any number $ (2^n)^2$ is not equal $ 2nCn$ what am I missing ?

I added an image for the question

@ Keen-ameteur He probably use $nCr$ to stand for $\binom{n}{r}$ — Ray, Sep 30 '20 at 08:26
It would help if you edited your query to express $\dfrac{n!}{k![(n-k)!]}$ by the mathJax $\binom{n}{k}$ . — user2661923, Sep 30 '20 at 08:27
With respect to this particular query, mathSE reviewers have responded (anyway), since you are new to the mathSE website. Going forward, in general however, it is generally better to use mathJax as much as possible. The https://math.stackexchange.com/help/notation link will get you started. — user2661923, Sep 30 '20 at 10:34
Also, as I indicated, mathSE reviewers are accustomed to the $\binom{n}{k}$ notation generated by the $\binom{n}{k}$ syntax. — user2661923, Sep 30 '20 at 10:36
Does this answer your question? Non-inductive, not combinatorial proof of $\sum_{i \mathop = 0}^n \binom n i^2 = \binom {2 n} n$ — StubbornAtom, Oct 01 '20 at 16:04

score 0 · Answer 1 · answered Sep 30 '20 at 08:28

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If your $nCk$ denotes $\displaystyle\binom nk$, the claim is obviously false, because the RHS is $2$.

answered Sep 30 '20 at 08:28

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The OP probably means $(2n)Cn$ on RHS. – drhab Sep 30 '20 at 08:31
yeah that what I meant – Mohamed shaddy Sep 30 '20 at 08:34
Sorry guys I made you confused it looks like my ministry of education in Egypt fked up the Binomial theorem – Mohamed shaddy Sep 30 '20 at 08:35
@Mohammedshaddy Just edit your question to repair (and use more decent terminology to show your frustrations). – drhab Sep 30 '20 at 08:37
well honestly it requires to learn Binomial theorem again from 0 from the internet because I actually don't see any problem in question, my books are the same for 50 years xD – Mohamed shaddy Sep 30 '20 at 08:41
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@Mohammedshaddy: you wrote that $C_k$ denotes a combination. This is wrong. $nCk$ does, be careful. – Sep 30 '20 at 08:43
I put an image . can u check it ? – Mohamed shaddy Sep 30 '20 at 09:09
@Mohammedshaddy: it absolutely confirms my previous comment. – Sep 30 '20 at 09:21

score 0 · Answer 2 · answered Sep 30 '20 at 09:55

I don't think the binomial theorem is useful here.

Suppose there are $2n$ persons and you want to select $n$ of them which can be done in $(2n)Cn$ ways.

Now we use a different method.

First you split up into two groups of $n$ persons each.

Then there are $nCk$ of selecting $k$ from the first group and $nC(n-k)$ of the other so there are $nCk\times nC(n-k)$ ways to select in such a way that $k$ persons of the first group are selected.

Then in total we find $\sum_{k=0}^nnCk\times nC(n-k)=\sum_{k=0}^n(nCk)^2$ selections.

Of course the two outcomes must be equal so:$$\sum_{k=0}^n(nCk)^2=(2n)Cn$$

Binomial theorem proof question

2 Answers2