# A proof.

1. Mar 4, 2004

### gimpy

Hi, im having a little trouble with this proof:

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

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

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

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

Am i on the right track?

2. Mar 4, 2004

Staff Emeritus
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.

3. Mar 5, 2004

### matt grime

you can proceed as follows:

note that $$\binom{n}{r}=\binom{n}{n-r}$$ 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).