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DrWillVKN
Mar13-11, 01:24 PM
1. The problem statement, all variables and given/known data
Show that if n is a positive integer, then C(n,0) < C(n,1) <....<C(n, floor(n/2)) = C(n, ceiling(n/2)) > .... > C(n, n)

2. Relevant equations

none

3. The attempt at a solution

Seems pretty direct, but I could be wrong. I suppose induction could work, too (or might be necessary).

The statement seems to imply that C(n, floor(n/2)) > C(n, k) for floor(n/2) > k >= 0 and C(n, ceiling(n/2)) > C(n, k) for ceiling(n/2) > k >= 0; n and k are positive integers. This can be stated because C(n, k) = C(n, n-k). P(n, k) = C(n, k)*k! by basic counting observation. Do these parts still need to be proven before they are shown to be equal? Does one need to use pascal's identity for this?

The trouble I have with the problem is with the flow of this proof. C(n, floor(n/2)) > C(n, k) for n>= k >= 0 because C(n, k) is greatest when k = n/2 for even, and k = (n+1)/2 = ceiling(n/2) and k = (n-1)/2 = floor(n/2) for odd, yet I don't know how I would go about stating this. It's too basic and intuitive.

Since C(n, k) = P(n, k)/k!, there are two cases to show:

1) When k is even, C(k, floor(k/2)) = k!/[(k/2)!(k/2)!], and C(k, ceiling(k/2)) = n!/[(k/2)!(k/2)!].
2) When k is odd, C(k, floor(k/2)) = k!/[(k-1/2)!(k+1/2)!] and C(k, ceiling(k/2)) = k!/[(k+1/2)!(k-1/2)!].

...

Thus, C(n, floor(n/2)) = C(n, ceiling(n/2)) > C(n, k) for n >=k >= 0 .

Dick
Mar13-11, 01:58 PM
I would try to write down an expression for the ratio C(n,k+1)/C(n,k). If it's greater than one C(n,k+1)>C(n,k), if it's less than one C(n,k+1)<C(n,k).

DrWillVKN
Mar13-11, 04:44 PM
So the ratio would fit everything on the left and right side in place, and then it comes together when k hits the floor and ceiling, and they're joined by equating them? I'm still confused about how to use the ratio (and for combinatorial proofs in general).

I have a hard time doing it by combinatorial proof, but by induction I have (for inductive step) the inductive hypothesis being the statement, adding the k + 1 term to the k term and showing pascal's identity, and then joining the left and right hand side by showing the ceiling and floor sets in the inductive step to be equal using algebra

Dick
Mar13-11, 04:48 PM
So the ratio would fit everything on the left and right side in place, and then it comes together when k hits the floor and ceiling, and they're joined by equating them? I'm still confused about how to use the ratio (and for combinatorial proofs in general).

I have a hard time doing it by combinatorial proof, but by induction I have (for inductive step) the inductive hypothesis being the statement, adding the k + 1 term to the k term and showing pascal's identity, and then joining the left and right hand side by showing the ceiling and floor sets in the inductive step to be equal using algebra.

I don't think induction is the easy way to do it. Did you figure out a simple expression for C(n,k+1)/C(n,k)? Just looking at it might help you organize your thoughts.