For a binary string of 6 elements with three ones and three zeros in each string, how do I calculate the total number of string permutations? For example, 3 possible strings would be 100011, 111000 and 101010.
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The number of binary strings with $m$ zeros and $n$ ones, is $\binom{n+m}{n}=\binom{n+m}{m}$. There are $n+m$ "positions," and you can choose $m$ of them to place the zeros; the ones will fill the rest.
angryavian
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Thanks for your reply. There are 3 zeros and 3 ones. Therefore, m=n=3. It is not clear to me how to apply the equality formula which you provided to determine the total number of strings which satisfy this specification? – Mel Dec 20 '16 at 06:26
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@Mel In your problem, $n = m = 3$. – N. F. Taussig Dec 20 '16 at 17:46
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@Mel If you are confused by the notation, this is what it means: $\binom{n+m}{n} = \binom{n+m}{m} = \frac{(n+m)!}{n! m!}$. – angryavian Dec 20 '16 at 22:14