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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...

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  1. ganeshie8
    • one year ago
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    suppose there are \(n\) men and \(n\) women and you want to choose a committee consisting of \(n\) people

  2. anonymous
    • one year ago
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    thats 2n choose n

  3. ganeshie8
    • one year ago
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    Yes, lets count it in an alternative way

  4. ganeshie8
    • one year ago
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    how many committees will be there with out women ?

  5. anonymous
    • one year ago
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    *

  6. anonymous
    • one year ago
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    1

  7. ganeshie8
    • one year ago
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    Yes, how many committees will be there with exactly 1 women ?

  8. anonymous
    • one year ago
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    n choose n-1

  9. ganeshie8
    • one year ago
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    try again

  10. anonymous
    • one year ago
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    n choose n-1 multiplied by n?

  11. ganeshie8
    • one year ago
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    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 \(\binom{n}{1}*\binom{n}{n-1}\)

  12. ganeshie8
    • one year ago
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    does that make sense

  13. anonymous
    • one year ago
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    yes, i get it

  14. anonymous
    • one year ago
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    now would i find the number of ways to make a committee with two women?

  15. ganeshie8
    • one year ago
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    yes find it, after that you will see the pattern

  16. anonymous
    • one year ago
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    alright, ill get back to you with the results :D

  17. ganeshie8
    • one year ago
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    take your time, we're almost done!

  18. anonymous
    • one year ago
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    hang on... n choose k and n choose (n-k) give the same result

  19. anonymous
    • one year ago
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    so the total possible combinations is just sums of the squares....

  20. ganeshie8
    • one year ago
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    Yep!

  21. anonymous
    • one year ago
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    but how do i equate it to 2n choose n?

  22. anonymous
    • one year ago
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    oh wait nevermind

  23. anonymous
    • one year ago
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    i get it

  24. ganeshie8
    • one year ago
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    good :) id like to see the complete proof if you don't mind

  25. anonymous
    • one year ago
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    sure :D

  26. ganeshie8
    • one year ago
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    take a screenshot and attach if psble

  27. anonymous
    • one year ago
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    i need to go eat dinner, i'll send it to you in about an hour, is that ok?

  28. ganeshie8
    • one year ago
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    take ur time

  29. anonymous
    • one year ago
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    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?

  30. ganeshie8
    • one year ago
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    sure that is also an interesting way to count first, may i see the previous proof...

  31. anonymous
    • one year ago
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    yes im just typing it up now

  32. anonymous
    • one year ago
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    sorry the codeisnt working...

  33. ganeshie8
    • one year ago
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    make this correction : \(k\le n\)

  34. ganeshie8
    • one year ago
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    other than that, the proof looks good!

  35. anonymous
    • one year ago
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    ok thanks!

  36. anonymous
    • one year ago
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    can you help me with the second part please?

  37. anonymous
    • one year ago
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    hello? are you still here @ganeshie8

  38. ganeshie8
    • one year ago
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    Hey!

  39. anonymous
    • one year ago
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    hi! can you please help me prove this with the "block walking"

  40. ganeshie8
    • one year ago
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    Consider a \(n\times n\)grid |dw:1440165928386:dw|

  41. anonymous
    • one year ago
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    right

  42. anonymous
    • one year ago
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    number of paths from bottom left to top right = \(\binom{2n}{n}\) |dw:1440168478785:dw|

  43. anonymous
    • one year ago
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    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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