i want to know how to do this question. please help
Show that the number of n-bit strings having exactly k 0's, with no two 0's consecutive, is C(n-k+1,k)
This answer is not yet complete. I'm going to go offline and take some time to think about the hard part.
Some hours later: still working. I'm going to leave the following in place for now, but here are some fresh notes:
-- begin additions
We really need to include the case for k=0 (1 string of n 1s, regardless of n).
I believe we need to examine the cases where k is at its maximum value.
Christian dating services in Northern Ireland::
I've tried various permutations but without success. Thank you. I trust that this answers your question. Do ask for clarification if it does not.
http://answers.google.com/answers/threadview?id=350365HOME |
What makes Coffee Gourmet::
Expires: 26 Oct 2005 15:47 PDT Question ID: 572989 enjoyment of a cup of coffee, the "coffee experience" in all it's permutations is pretty subjective.
http://answers.google.com/answers/threadview/id/572989.htmlHOME |
If n is odd, the maximum value for k is (n+1)/2. This allows 1 string: alternating 0s and 1s, starting with 0. (As n-k+1=k, the formula holds.)
If n is even, the maximum value for k is n/2. There are k+1 possible strings: each consists of m consecutive 01 pairs, followed by k-m consecutive 10 pairs, where 0≤m≤k.
I now suspect the proof should proceed in a similar manner to that for the terms in Pascal's triangle. Calling the function in question Z(n,k), show a derivation of
Algorithm for lowest bid path::
The question is then, what do you want to minimize when you mean "lowest bid .. The reduction in the number of permutations was relatively minor with the
http://answers.google.com/answers/threadview/id/752652.htmlHOME |
Z(n+1,k+1)
from Z(n,k) and Z(n,k+1)
by a mapping of the possible strings.
So I'm now playing with the binary strings in hopes of finding a good expression of that mapping.
-- end additions; the rest is what I threw out at first
First, we note that the expression
C(n-k+1, k) is meaningless unless
n+1 ≥ 2k
But this makes sense, because we are looking for strings with no consecutive 0's, and therefore there has to be room for at least k-1 1's in each string.
Second, let us note that if k=1, then we are looking for the number of different binary strings of length n consisting of one zero and (n-1) ones, and obviously there are n positions for the single zero. Since
Large Factorial::
Request for Question Clarification by richard-ga on 15 Apr 2005 09:32 PDT . of summarizing the number of permutations possible given a specific set of
http://answers.google.com/answers/threadview/id/509662.htmlHOME |
C(n-1+1,1) = n
the formula holds for any n as long as k=1.
From here, I think we proceed by mathematical induction. That is, we suppose the formula holds for some n and k such that
n+1>2k
and consider the case of an n-bit string and k+1 zeros.
What we need is a way to identify how many ones, in each string we had for the case of n and k, we could replace with another zero. That is the hard part.
Pre-Article:What happend on raw last night? 10 for the best answer?
Next-Article:Is there a reason for everything?
|
Permutations Question?