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A proof.

  1. Mar 4, 2004 #1
    Hi, im having a little trouble with this proof:

    Let n be a positive integer. What is the largest binomial coefficient [tex]C(n,r)[/tex] where r is a nonnegative integer less than or equal to n? Prove your answer is correct.

    So let [tex]r = \lfloor{\frac{n}{2}\rfloor} [/tex] or [tex]r = \lceil{\frac{n}{2}\rceil} [/tex] then [tex]\left( \begin{array}{c} n \\ r \end{array} \right)[/tex] is the largest binomial coefficient.

    Now im having trouble with the proof. Where do i begin?
    Maybe something like this?

    [tex]\left( \begin{array}{c} n \\ \lfloor \frac{n}{2} \rfloor \end{array} \right) = \frac{n!}{\left(\lfloor \frac{n}{2} \rfloor \right)! \left(n - \lfloor \frac{n}{2} \rfloor \right)!} = ......[/tex]

    Am i on the right track?
  2. jcsd
  3. Mar 4, 2004 #2


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    Yes you are on the right track. There are two cases depending on whether n is odd or even. Obviously the middle values are the biggest, just think of the Pascal triangle. Then I suppose you would prove that (in the even case) C(n,n/2) > C(n,r) for r < n and r not equal n/2. Consider writing n as 2m, so n/2 = m, and splitting out the facors: C(n,m) = n(n-1)...(m+1)/m(m-1)...1 compared to C(n,r) = n(n-1)...r/(n-r)(n-r-1)...1, and match factors.
  4. Mar 5, 2004 #3

    matt grime

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    you can proceed as follows:

    note that [tex]\binom{n}{r}=\binom{n}{n-r}[/tex] and that it thus suffices to show that the coeffs for r=0 to [n/2] (integral part) are increasing.

    This follows from a *careful* induction argument (it's best to have two base cases going for n=2 and 3 say).
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