Mixing
Counting binary integers with a given number of zeroes
$begingroup$
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
edited Jan 19 at 20:04
Did
248k23224463
asked Jan 19 at 20:00
varunrvarunr
1
$endgroup$
add a comment |
$begingroup$
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
edited Jan 19 at 20:04
Did
248k23224463
asked Jan 19 at 20:00
varunrvarunr
1
$endgroup$
$begingroup$
Because the number 0100 is counted as having two zeroes, not three.
$endgroup$
– Did
Jan 19 at 20:03
add a comment |
$begingroup$
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
edited Jan 19 at 20:04
Did
248k23224463
asked Jan 19 at 20:00
varunrvarunr
1
$endgroup$
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
edited Jan 19 at 20:04
Did
248k23224463
asked Jan 19 at 20:00
varunrvarunr
1
edited Jan 19 at 20:04
Did
248k23224463
asked Jan 19 at 20:00
varunrvarunr
1
edited Jan 19 at 20:04
Did
248k23224463
edited Jan 19 at 20:04
Did
248k23224463
edited Jan 19 at 20:04
Did
248k23224463
248k23224463
asked Jan 19 at 20:00
varunrvarunr
1
asked Jan 19 at 20:00
varunrvarunr
1
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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
answered Jan 19 at 21:47
Daniel MathiasDaniel Mathias
1,31518
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
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votes
$begingroup$
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
answered Jan 19 at 21:47
Daniel MathiasDaniel Mathias
1,31518
$endgroup$
add a comment |
$begingroup$
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
answered Jan 19 at 21:47
Daniel MathiasDaniel Mathias
1,31518
$endgroup$
add a comment |
$begingroup$
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
answered Jan 19 at 21:47
Daniel MathiasDaniel Mathias
1,31518
$endgroup$
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
answered Jan 19 at 21:47
Daniel MathiasDaniel Mathias
1,31518
answered Jan 19 at 21:47
Daniel MathiasDaniel Mathias
1,31518
answered Jan 19 at 21:47
Daniel MathiasDaniel Mathias
1,31518
answered Jan 19 at 21:47
Daniel MathiasDaniel Mathias
1,31518
1,31518
add a comment |
add a comment |
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$begingroup$
Because the number 0100 is counted as having two zeroes, not three.
$endgroup$
– Did
Jan 19 at 20:03