Re: Probabilities problem (Easy-looking but tough for me).

From: George (gtsavdar_at_auth.gr)
Date: 09/14/04 

Date: 14 Sep 2004 09:09:16 -0700

"Jose H. Nieto" <jhnieto@yahoo.com> wrote in message news:<2qmkcrF11fconU1@uni-berlin.de>...
> > X1)Let the function f(A,B), to be the number of all different
> > arrangements of the 2 numbers (0 and 1), when these 2 numbers(0,1) are
> > placed in A positions and when in every arrangement we should have at
> > least B 1 placed side by side.
> >
> > For example it's true that f(5,3)=8, since when we have 5 places, all
> > the different arrangements of 0 and 1, that contain at least three
> > 1(ones) placed side by side are the following:
> >
> > 11100
> > 11101
> > 11110
> > 11111
> > 00111
> > 10111
> > 01111
> > 01110
> >
> > The problem is to find the form of the function f(A,B).
>
> f(A,B) = (A-B+2)*2^(A-B-1)
>
> (Hint: count the number of arrangements such that the first run of B 1's
> begins at position k, for k=1,...,A-B+1, then sum).
>

Ooops, sorry for the little mistake on the equation. I was having in
my mind an example with B=2 and instead of writing B i wrote 2 at the
equation:

So here is the right equation:

f(A,B) = (A-B+2)*2^(A-B-1) - (f(1,B) + f(2,B) + f(3,B) +...+
f(A-B-1,B))

where f(x,B)=0 if x<B.

But even this way, i don't see how i can find the function f(A,B) with
this.

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**Your mistake is that when you count the number of arrangements at
position 7 for example, for the f(14,3)----------->

_ _ _ _ _ 0111 _ _ _ _ _ the number of all possible arrangements is
not 2^(14-3-1) as it is proposed by your solution but it's smaller,
since the arrangement:
0111_ 0111 _ _ _ _ _ is already calculated at position 2.

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