How many strings of n bits are there which differ by exactly m bits?
$begingroup$
For example:
If n = 8 and m = 0: ans = 1
If n = 8 and m = 1: ans = 8
If n = 8 and m = 2: ans = ?
If n =256 and m = 3: ans = ?
linear-algebra combinatorics bit-strings
New contributor
$endgroup$
add a comment |
$begingroup$
For example:
If n = 8 and m = 0: ans = 1
If n = 8 and m = 1: ans = 8
If n = 8 and m = 2: ans = ?
If n =256 and m = 3: ans = ?
linear-algebra combinatorics bit-strings
New contributor
$endgroup$
$begingroup$
Isn't this just $nchoose m$? Perhaps I am misinterpreting the question..
$endgroup$
– John Doe
Jan 7 at 19:30
add a comment |
$begingroup$
For example:
If n = 8 and m = 0: ans = 1
If n = 8 and m = 1: ans = 8
If n = 8 and m = 2: ans = ?
If n =256 and m = 3: ans = ?
linear-algebra combinatorics bit-strings
New contributor
$endgroup$
For example:
If n = 8 and m = 0: ans = 1
If n = 8 and m = 1: ans = 8
If n = 8 and m = 2: ans = ?
If n =256 and m = 3: ans = ?
linear-algebra combinatorics bit-strings
linear-algebra combinatorics bit-strings
New contributor
New contributor
New contributor
asked Jan 7 at 19:28
d eid ei
31
31
New contributor
New contributor
$begingroup$
Isn't this just $nchoose m$? Perhaps I am misinterpreting the question..
$endgroup$
– John Doe
Jan 7 at 19:30
add a comment |
$begingroup$
Isn't this just $nchoose m$? Perhaps I am misinterpreting the question..
$endgroup$
– John Doe
Jan 7 at 19:30
$begingroup$
Isn't this just $nchoose m$? Perhaps I am misinterpreting the question..
$endgroup$
– John Doe
Jan 7 at 19:30
$begingroup$
Isn't this just $nchoose m$? Perhaps I am misinterpreting the question..
$endgroup$
– John Doe
Jan 7 at 19:30
add a comment |
1 Answer
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$begingroup$
Just choose the $m$ bits out of $n$ to be different, so $n choose m$
$endgroup$
$begingroup$
You're right of course. Thanks!
$endgroup$
– d ei
Jan 9 at 20:50
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
Just choose the $m$ bits out of $n$ to be different, so $n choose m$
$endgroup$
$begingroup$
You're right of course. Thanks!
$endgroup$
– d ei
Jan 9 at 20:50
add a comment |
$begingroup$
Just choose the $m$ bits out of $n$ to be different, so $n choose m$
$endgroup$
$begingroup$
You're right of course. Thanks!
$endgroup$
– d ei
Jan 9 at 20:50
add a comment |
$begingroup$
Just choose the $m$ bits out of $n$ to be different, so $n choose m$
$endgroup$
Just choose the $m$ bits out of $n$ to be different, so $n choose m$
edited Jan 7 at 19:31
John Doe
11.1k11238
11.1k11238
answered Jan 7 at 19:30
Ross MillikanRoss Millikan
293k23197372
293k23197372
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You're right of course. Thanks!
$endgroup$
– d ei
Jan 9 at 20:50
add a comment |
$begingroup$
You're right of course. Thanks!
$endgroup$
– d ei
Jan 9 at 20:50
$begingroup$
You're right of course. Thanks!
$endgroup$
– d ei
Jan 9 at 20:50
$begingroup$
You're right of course. Thanks!
$endgroup$
– d ei
Jan 9 at 20:50
add a comment |
d ei is a new contributor. Be nice, and check out our Code of Conduct.
d ei is a new contributor. Be nice, and check out our Code of Conduct.
d ei is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
Isn't this just $nchoose m$? Perhaps I am misinterpreting the question..
$endgroup$
– John Doe
Jan 7 at 19:30