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anonymous
 one year ago
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
 one year ago
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...

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ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6suppose there are \(n\) men and \(n\) women and you want to choose a committee consisting of \(n\) people

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6Yes, lets count it in an alternative way

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6how many committees will be there with out women ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6Yes, how many committees will be there with exactly 1 women ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0n choose n1 multiplied by n?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6you can choose \(1\) women from the group of \(n\) women in \(\binom{n}{1}\) ways after that, the remaining \(n1\) men can be chosen from the group of \(n\) men in \(\binom{n}{n1}\) ways so total \(n\) member committees with exactly \(1\) women is \(\binom{n}{1}*\binom{n}{n1}\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6does that make sense

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now would i find the number of ways to make a committee with two women?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6yes find it, after that you will see the pattern

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0alright, ill get back to you with the results :D

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6take your time, we're almost done!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hang on... n choose k and n choose (nk) give the same result

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so the total possible combinations is just sums of the squares....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but how do i equate it to 2n choose n?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6good :) id like to see the complete proof if you don't mind

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6take a screenshot and attach if psble

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i need to go eat dinner, i'll send it to you in about an hour, is that ok?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i 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
 one year ago
Best ResponseYou've already chosen the best response.6sure that is also an interesting way to count first, may i see the previous proof...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes im just typing it up now

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sorry the codeisnt working...

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6make this correction : \(k\le n\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6other than that, the proof looks good!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can you help me with the second part please?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hello? are you still here @ganeshie8

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hi! can you please help me prove this with the "block walking"

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6Consider a \(n\times n\)grid dw:1440165928386:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0number of paths from bottom left to top right = \(\binom{2n}{n}\) dw:1440168478785:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0suppose your friend's home is located at \((k,~nk)\), where \(k\in \left\{0,1,2,\ldots ,n \right\}\). then number of paths through \((k,nk)\) is given by \(\binom{n}{k}*\binom{n}{nk} = \binom{n}{k}^2\) ..... https://proofwiki.org/wiki/Sum_of_Squares_of_Binomial_Coefficients
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