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Probability quiz

  1. Apr 18, 2013 #1
    In the expansion of ((x1 + x2 +.... + x33) ^ 4) how many monomials are of the form (xi ^ 2)*(xj^2)
    with i not equal with j

    if we add the coefficients of all these mononymon what is the sum?

    E.g. the expansion of (x1 + x2 + x3) ^ 4 the requested number is 18

    this is a probability quiz but i can think how to solve it with theory of probabilities any ideas for howmto beggin to solve this?
     
  2. jcsd
  3. Apr 18, 2013 #2

    mfb

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    There is nothing probabilistic here, but combinatorics is important.
    Have a closer look at your example - you should see some pattern about the coefficients. Find that pattern, and extend it to 33.
     
  4. Apr 18, 2013 #3

    haruspex

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    I don't think so. I believe 6x12x22 constitutes one monomial.
     
  5. Apr 19, 2013 #4

    Bacle2

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    I think it comes down to counting the number of ways of selecting a pair from {1,2,....,33}. For each

    pair (xi[/SUBi,xj)) ,there will be one term xi[/SUP]2[/SUP]*xj2 . The general binomial coefficient ci in

    ci x1y1x2y2x3y3 x4y4 counts

    precisely the number of ways of selecting a total of y_1 x_1's, y_2 x_2's, etc. in the product.

    Then ,for

    xi2xj2, you're counting the number of ways of selecting exactly 2 x_i's and 2 x_j's from the expansion :

    (x1+x2+x3+x4)...... ( 4 times )

    How many ways can you choose a pair (xi,xj) from

    (x1,x2, x3,...,x33)?
     
    Last edited: Apr 19, 2013
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