I need help to prove that $$\binom{n}{0}^2 + \binom{n}{1}^2 + \cdots + \binom{n}{n}^2 = \binom{2n}{n}.$$ using committee forming...

- anonymous

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- ganeshie8

suppose there are \(n\) men and \(n\) women
and you want to choose a committee consisting of \(n\) people

- anonymous

thats 2n choose n

- ganeshie8

Yes, lets count it in an alternative way

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## More answers

- ganeshie8

how many committees will be there with out women ?

- anonymous

*

- anonymous

1

- ganeshie8

Yes, how many committees will be there with exactly 1 women ?

- anonymous

n choose n-1

- ganeshie8

try again

- anonymous

n choose n-1 multiplied by n?

- ganeshie8

you can choose \(1\) women from the group of \(n\) women in \(\binom{n}{1}\) ways
after that, the remaining \(n-1\) men can be chosen from the group of \(n\) men in \(\binom{n}{n-1}\) ways
so total \(n\) member committees with exactly \(1\) women is \(\binom{n}{1}*\binom{n}{n-1}\)

- ganeshie8

does that make sense

- anonymous

yes, i get it

- anonymous

now would i find the number of ways to make a committee with two women?

- ganeshie8

yes find it, after that you will see the pattern

- anonymous

alright, ill get back to you with the results :D

- ganeshie8

take your time, we're almost done!

- anonymous

hang on... n choose k and n choose (n-k) give the same result

- anonymous

so the total possible combinations is just sums of the squares....

- ganeshie8

Yep!

- anonymous

but how do i equate it to 2n choose n?

- anonymous

oh wait nevermind

- anonymous

i get it

- ganeshie8

good :) id like to see the complete proof if you don't mind

- anonymous

sure :D

- ganeshie8

take a screenshot and attach if psble

- anonymous

i need to go eat dinner, i'll send it to you in about an hour, is that ok?

- ganeshie8

take ur time

- anonymous

i also need to prove the same thing using the "block walking" method... but i dont know how. Do you think you can try to help me with this too? please?

- ganeshie8

sure that is also an interesting way to count
first, may i see the previous proof...

- anonymous

yes im just typing it up now

- anonymous

sorry the codeisnt working...

- ganeshie8

make this correction : \(k\le n\)

- ganeshie8

other than that, the proof looks good!

- anonymous

ok
thanks!

- anonymous

can you help me with the second part please?

- anonymous

hello? are you still here @ganeshie8

- ganeshie8

Hey!

- anonymous

hi! can you please help me prove this with the "block walking"

- ganeshie8

Consider a \(n\times n\)grid
|dw:1440165928386:dw|

- anonymous

right

- anonymous

number of paths from bottom left to top right = \(\binom{2n}{n}\)
|dw:1440168478785:dw|

- anonymous

suppose your friend's home is located at \((k,~n-k)\), where \(k\in \left\{0,1,2,\ldots ,n \right\}\).
then number of paths through \((k,n-k)\) is given by \(\binom{n}{k}*\binom{n}{n-k} = \binom{n}{k}^2\)
.....
https://proofwiki.org/wiki/Sum_of_Squares_of_Binomial_Coefficients

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