anonymous
  • anonymous
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...
Mathematics
schrodinger
  • schrodinger
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ganeshie8
  • ganeshie8
suppose there are \(n\) men and \(n\) women and you want to choose a committee consisting of \(n\) people
anonymous
  • anonymous
thats 2n choose n
ganeshie8
  • ganeshie8
Yes, lets count it in an alternative way

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ganeshie8
  • ganeshie8
how many committees will be there with out women ?
anonymous
  • anonymous
*
anonymous
  • anonymous
1
ganeshie8
  • ganeshie8
Yes, how many committees will be there with exactly 1 women ?
anonymous
  • anonymous
n choose n-1
ganeshie8
  • ganeshie8
try again
anonymous
  • anonymous
n choose n-1 multiplied by n?
ganeshie8
  • 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
  • ganeshie8
does that make sense
anonymous
  • anonymous
yes, i get it
anonymous
  • anonymous
now would i find the number of ways to make a committee with two women?
ganeshie8
  • ganeshie8
yes find it, after that you will see the pattern
anonymous
  • anonymous
alright, ill get back to you with the results :D
ganeshie8
  • ganeshie8
take your time, we're almost done!
anonymous
  • anonymous
hang on... n choose k and n choose (n-k) give the same result
anonymous
  • anonymous
so the total possible combinations is just sums of the squares....
ganeshie8
  • ganeshie8
Yep!
anonymous
  • anonymous
but how do i equate it to 2n choose n?
anonymous
  • anonymous
oh wait nevermind
anonymous
  • anonymous
i get it
ganeshie8
  • ganeshie8
good :) id like to see the complete proof if you don't mind
anonymous
  • anonymous
sure :D
ganeshie8
  • ganeshie8
take a screenshot and attach if psble
anonymous
  • anonymous
i need to go eat dinner, i'll send it to you in about an hour, is that ok?
ganeshie8
  • ganeshie8
take ur time
anonymous
  • 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
  • ganeshie8
sure that is also an interesting way to count first, may i see the previous proof...
anonymous
  • anonymous
yes im just typing it up now
anonymous
  • anonymous
sorry the codeisnt working...
ganeshie8
  • ganeshie8
make this correction : \(k\le n\)
ganeshie8
  • ganeshie8
other than that, the proof looks good!
anonymous
  • anonymous
ok thanks!
anonymous
  • anonymous
can you help me with the second part please?
anonymous
  • anonymous
hello? are you still here @ganeshie8
ganeshie8
  • ganeshie8
Hey!
anonymous
  • anonymous
hi! can you please help me prove this with the "block walking"
ganeshie8
  • ganeshie8
Consider a \(n\times n\)grid |dw:1440165928386:dw|
anonymous
  • anonymous
right
anonymous
  • anonymous
number of paths from bottom left to top right = \(\binom{2n}{n}\) |dw:1440168478785:dw|
anonymous
  • 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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