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I have a set of $26,000$ values. Each value has the option of being $1$ or $0$. How do I calculate the number of potential combinations of $1$'s and $0$'s that exist for $26,000$ values?

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  • $\begingroup$ Depends on whether each of the 26,000 places are unique ( for eg. numbered ) or identical. Permutation in the first case and combination in the second. $\endgroup$ Jun 23, 2015 at 16:33
  • $\begingroup$ Consider you problem. You have $26000$ indipendent variables each one can assume either the value $0$ or the value $1$. A solution for your problem should associate to every variable either the value $0$ or the value $1$.... $\endgroup$ Jun 23, 2015 at 16:34
  • $\begingroup$ The values are identical (either a 1 or a 0), not unique. If the first value in the set has a value of "1", it is precisely the same value as if the 24,000th value in the set had a value of "1". $\endgroup$
    – nicktendo
    Jun 23, 2015 at 16:40
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    $\begingroup$ @nicktendo So it's simple. You can have all zeroes or all ones, and the in between cases (one one, two ones, three ones...), i.e. there are a total of 26,001 cases. $\endgroup$ Jun 23, 2015 at 16:47

2 Answers 2

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If you are interested in ordering the values (e.g. $101 \neq 110 $) then you have a permutation of the $n=2$ elements ($0$ and $1$) and the length of these sequences is $k=26000$. So you can use the formula $n^k=2^{26000}$

If order doesn't matter (e.g. $101=110$ because there are the same numbers of $0$ and $1$), you have a combination of your elements $0$ and $1$. In this case you need the formula $${n+k-1 \choose k}={2+26000-1 \choose 26000}=\frac{26001!}{26000!}=26001$$

You can find the explanations here http://redooc.com/it/matematica-statistica/calcolo-combinatorio/disposizioni for permutations (in italian they're called "disposizioni") or here for combinations

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  • $\begingroup$ Obviously, the last formula gives 26001 not 26000. $\endgroup$
    – Leo
    Jun 23, 2015 at 22:27
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If you had $2$ values instead of $26,000$, the possibilities would be $$00,01,10,11$$

If there were three values, you'd have the four former possibilities for the two first digits, times the two possibilities for the last one. This makes: $$000,001,010,011,100,101,110,111$$ that is, $4\times 2=8$ possibilities. Can you extrapolate?

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  • $\begingroup$ I'm looking for a formula to solve my problem, not the basic logic from which it would be extrapolated. $\endgroup$
    – nicktendo
    Jun 23, 2015 at 16:43

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