Counting binary integers with a given number of zeroes












0












$begingroup$


Problem:
Let n be a positive integer. A number is randomly chosen from 1 to 2^n. The probability that it has exactly 6 zeroes in its binary representation is 1989/16384. What is n?



Solution:
For n≥7, the number of integers from 1 up to 2^n that has exactly 6 zeroes in its binary representation is (nC7). This can be seen by observing that among n candidates (for places of 0's and 1's), we need to select 7 places, 1 for the leading digit 1 and 6 more after that for the 6 zeroes. So, we get (nC7)/2^n=1989/16384. By looking at the denominators, we notice that n≥14. Since 1989 is a multiple of 9, we get n=15 as the smallest possible n. But checking this, we see it doesn't work. The next smallest n is n=18, which works since (187)=9×13×16×17=1989×16. So n=18.



The correct answer is: 18



My Question:
I don't understand why you have to do nC7 and not nC6 since you are choosing 6 zeroes.






combinatorics binary







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$endgroup$












  • $begingroup$
    Because the number 0100 is counted as having two zeroes, not three.
    $endgroup$
    – Did
    Jan 19 at 20:03
















0












$begingroup$


Problem:
Let n be a positive integer. A number is randomly chosen from 1 to 2^n. The probability that it has exactly 6 zeroes in its binary representation is 1989/16384. What is n?



Solution:
For n≥7, the number of integers from 1 up to 2^n that has exactly 6 zeroes in its binary representation is (nC7). This can be seen by observing that among n candidates (for places of 0's and 1's), we need to select 7 places, 1 for the leading digit 1 and 6 more after that for the 6 zeroes. So, we get (nC7)/2^n=1989/16384. By looking at the denominators, we notice that n≥14. Since 1989 is a multiple of 9, we get n=15 as the smallest possible n. But checking this, we see it doesn't work. The next smallest n is n=18, which works since (187)=9×13×16×17=1989×16. So n=18.



The correct answer is: 18



My Question:
I don't understand why you have to do nC7 and not nC6 since you are choosing 6 zeroes.






combinatorics binary







share|cite|improve this question











$endgroup$












  • $begingroup$
    Because the number 0100 is counted as having two zeroes, not three.
    $endgroup$
    – Did
    Jan 19 at 20:03














0












0








0





$begingroup$


Problem:
Let n be a positive integer. A number is randomly chosen from 1 to 2^n. The probability that it has exactly 6 zeroes in its binary representation is 1989/16384. What is n?



Solution:
For n≥7, the number of integers from 1 up to 2^n that has exactly 6 zeroes in its binary representation is (nC7). This can be seen by observing that among n candidates (for places of 0's and 1's), we need to select 7 places, 1 for the leading digit 1 and 6 more after that for the 6 zeroes. So, we get (nC7)/2^n=1989/16384. By looking at the denominators, we notice that n≥14. Since 1989 is a multiple of 9, we get n=15 as the smallest possible n. But checking this, we see it doesn't work. The next smallest n is n=18, which works since (187)=9×13×16×17=1989×16. So n=18.



The correct answer is: 18



My Question:
I don't understand why you have to do nC7 and not nC6 since you are choosing 6 zeroes.






combinatorics binary







share|cite|improve this question











$endgroup$




Problem:
Let n be a positive integer. A number is randomly chosen from 1 to 2^n. The probability that it has exactly 6 zeroes in its binary representation is 1989/16384. What is n?



Solution:
For n≥7, the number of integers from 1 up to 2^n that has exactly 6 zeroes in its binary representation is (nC7). This can be seen by observing that among n candidates (for places of 0's and 1's), we need to select 7 places, 1 for the leading digit 1 and 6 more after that for the 6 zeroes. So, we get (nC7)/2^n=1989/16384. By looking at the denominators, we notice that n≥14. Since 1989 is a multiple of 9, we get n=15 as the smallest possible n. But checking this, we see it doesn't work. The next smallest n is n=18, which works since (187)=9×13×16×17=1989×16. So n=18.



The correct answer is: 18



My Question:
I don't understand why you have to do nC7 and not nC6 since you are choosing 6 zeroes.






combinatorics binary




combinatorics binary






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 19 at 20:04









Did

248k23224463




248k23224463










asked Jan 19 at 20:00









varunrvarunr

1




1












  • $begingroup$
    Because the number 0100 is counted as having two zeroes, not three.
    $endgroup$
    – Did
    Jan 19 at 20:03


















  • $begingroup$
    Because the number 0100 is counted as having two zeroes, not three.
    $endgroup$
    – Did
    Jan 19 at 20:03
















$begingroup$
Because the number 0100 is counted as having two zeroes, not three.
$endgroup$
– Did
Jan 19 at 20:03




$begingroup$
Because the number 0100 is counted as having two zeroes, not three.
$endgroup$
– Did
Jan 19 at 20:03










1 Answer
1






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oldest

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0












$begingroup$

Starting with $n$ potential bits:



oooooooooooooooooo



Choose $7$ bits: (You are choosing $6$ bits to be $0$s and $1$ bit to be the leftmost $1$)



oooooooooooooooooo

  ^ ^^ ^ ^ ^   ^


The leftmost will be a $1$, the other $6$ will be $0$s



oo1o00o0o0o0ooo0o



Of the remaining bits, those to the left of the $1$ will be dropped, and those to the right will be $1$s



..110010101011101






share|cite|improve this answer









$endgroup$













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    1 Answer
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    1 Answer
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    0












    $begingroup$

    Starting with $n$ potential bits:



    oooooooooooooooooo



    Choose $7$ bits: (You are choosing $6$ bits to be $0$s and $1$ bit to be the leftmost $1$)



    oooooooooooooooooo
    
  ^ ^^ ^ ^ ^   ^


    The leftmost will be a $1$, the other $6$ will be $0$s



    oo1o00o0o0o0ooo0o



    Of the remaining bits, those to the left of the $1$ will be dropped, and those to the right will be $1$s



    ..110010101011101






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Starting with $n$ potential bits:



      oooooooooooooooooo



      Choose $7$ bits: (You are choosing $6$ bits to be $0$s and $1$ bit to be the leftmost $1$)



      oooooooooooooooooo
      
  ^ ^^ ^ ^ ^   ^


      The leftmost will be a $1$, the other $6$ will be $0$s



      oo1o00o0o0o0ooo0o



      Of the remaining bits, those to the left of the $1$ will be dropped, and those to the right will be $1$s



      ..110010101011101






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Starting with $n$ potential bits:



        oooooooooooooooooo



        Choose $7$ bits: (You are choosing $6$ bits to be $0$s and $1$ bit to be the leftmost $1$)



        oooooooooooooooooo
        
  ^ ^^ ^ ^ ^   ^


        The leftmost will be a $1$, the other $6$ will be $0$s



        oo1o00o0o0o0ooo0o



        Of the remaining bits, those to the left of the $1$ will be dropped, and those to the right will be $1$s



        ..110010101011101






        share|cite|improve this answer









        $endgroup$



        Starting with $n$ potential bits:



        oooooooooooooooooo



        Choose $7$ bits: (You are choosing $6$ bits to be $0$s and $1$ bit to be the leftmost $1$)



        oooooooooooooooooo
        
  ^ ^^ ^ ^ ^   ^


        The leftmost will be a $1$, the other $6$ will be $0$s



        oo1o00o0o0o0ooo0o



        Of the remaining bits, those to the left of the $1$ will be dropped, and those to the right will be $1$s



        ..110010101011101







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 19 at 21:47









        Daniel MathiasDaniel Mathias

        1,31518




        1,31518






























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