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DrWillVKN
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Show that if n is a positive integer, then C(n,0) < C(n,1) <...<C(n, floor(n/2)) = C
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)
none
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 .
Homework Statement
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)
Homework Equations
none
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 .
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