If you have 6 0's and 5 1's, like in binary, how many different ways can you order them?

Also, there is a popup saying that the question appears subjective. Is it?

  • $\begingroup$ $\frac{(6+5)!}{6!\times5!}$ $\endgroup$ – barak manos May 26 '16 at 8:41
  • $\begingroup$ @barakmanos coould you explain this? $\endgroup$ – SuperNinja741 Does Gaming May 26 '16 at 8:47


You have 11 places to fill, and you can put five ones in the eleven places, the rest will be zeroes. Once you choose (hint, hint) the positions of the ones, the zeroes are fixed. Each choice of ones will give exactly one unique ordering.


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