PRODUCING BINARY NUMBERS

[PINKFLOYD]

Good Morning from Greece. I would very much appreciate if you could give a hand on this one:
Let it be a binary number with r digits produced by k "aces" and r-k "zeroes". How many such defferent binaries can be produced? And further more what is the possibility for a randomly selected "zero" digit of the binary to be next to an "ace" (it doesnt matter if the "ace" is left or right to the "zero")
Now , for small k and r , we can answer by constructing a tree or for greater numbers with just plug them in to a computer but i was wondering if there is a analytical solution to the problem
Thank you for your time in advance.

 

[pka]

Good Morning from Greece. I would very much appreciate if you could give a hand on this one: Let it be a binary number with r digits produced by k "aces" and r-k "zeroes". How many such defferent binaries can be produced? And further more what is the possibility for a randomly selected "zero" digit of the binary to be next to an "ace" (it doesnt matter if the "ace" is left or right to the "zero")

Strings composed of zeros and ones are are known as binary strings (bit-strings).
A bit-string of length \(r\), with \(k\) ones and \(r-k\) zeros can be arranged in \(\dfrac{r!}{(k!)(r-k)!}\) ways.
Frankly, the phrase "possibility for a randomly selected "zero" concerns me.

 

[Cubist]

We are on the dark side of the moon ?. A specific example may help to clarify this question. If we have the following binary number...

Binary number:  0 1 1 0 0 0 1 0
Bit position:   7 6 5 4 3 2 1 0

then we can list all the bit positions that contain 0 and we get { 0,2,3,4,7 }

The probability that one of these bit positions will be adjacent to a 1 in this binary number is 4/5.

NOW @PINKFLOYD might be asking what the probability would be if we follow this procedure on a random binary number (that has k ones and r-k zeroes). Is this correct?

 

[PINKFLOYD]

Thanks for replying so quickly. Sorry about my spelling mistakes. It has been a while since I've written in english. In Greece we often use the word ace instead of one because in greek language the plural of the word one is not widely common. We could use the word unit instead but we kind of save it for metric system of frame reference. I also share your concerns about the second part of the problem but it was given to me in that particular way. I think it means this. Let's suppose that from all the bit strings that are produced we select only those which have the x-th digit a zero where x is a number between 1 and r. Next to that zero there can be either one or zero. Now i think the possibility is not 50% because we have limited resources of zeros and ones and we must consider how many of them have already been used till we have reached the x digit.What is that probability . Frankly ocant think a beter explanation.

 

[PINKFLOYD]

We are on the dark side of the moon ?. A specific example may help to clarify this question. If we have the following binary number...

Binary number:  0 1 1 0 0 0 1 0
Bit position:   7 6 5 4 3 2 1 0

then we can list all the bit positions that contain 0 and we get { 0,2,3,4,7 }

The probability that one of these bit positions will be adjacent to a 1 in this binary number is 4/5.

NOW @PINKFLOYD might be asking what the probability would be if we follow this procedure on a random binary number (that has k ones and r-k zeroes). Is this correct?

....that was also my idea at first but on second thought we also must consider the position of the zero . That means we have in someway to combine two possibilities. For example what is the possibility for the 3rd bit to be zero and at the same time next to that zero is a one. Thank you .

 

[pka]

I think it means this. Let's suppose that from all the bit strings that are produced we select only those which have the x-th digit a zero where x is a number between 1 and r. Next to that zero there can be either one or zero. Now i think the possibility is not 50% because we have limited resources of zeros and ones and we must consider how many of them have already been used till we have reached the x digit.What is that probability .

That is still hard to parses. If there are eight ones and ten zeros then there \(\dfrac{18!}{8!\cdot 10!}\) possible bit-strings.
Of those there are \(\dfrac{17!}{8!\cdot 9!}\) that have a zero in the tenth position. What is the probability that there is a one in the ninth or the eleventh position. @PINKFLOYD is that how you understand the meaning?

 

[PINKFLOYD]

That is still hard to parses. If there are eight ones and ten zeros then there \(\dfrac{18!}{8!\cdot 10!}\) possible bit-strings.
Of those there are \(\dfrac{17!}{8!\cdot 9!}\) that have a zero in the tenth position. What is the probability that there is a one in the ninth or the eleventh position. @PINKFLOYD is that how you understand the meaning?

...yes thats about it. Thank you.

 

[PINKFLOYD]

Please check the attached file.