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  • $\begingroup$ In case of your matrix, Plus@@Flatten[binarym] should work. There may be faster ways using Trace or similar. Please note that your outer round parentheses do not actually do anything. $\endgroup$ Commented Jan 2, 2018 at 3:12
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    $\begingroup$ Count can take a level specification (say 2, for a matrix): Count[binarym, 0, 2], while something like Total[1 - binarym, 2] might be faster in some cases. $\endgroup$ Commented Jan 2, 2018 at 3:12
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    $\begingroup$ Select all the zeros and then find out how many of them there are: Length@Select[Flatten[binarym], # == 0 &] This gives 17, which is the expected answer. $\endgroup$ Commented Jan 2, 2018 at 3:14
  • $\begingroup$ Times @@ Dimensions[binarym] - Total@Flatten@binarym $\endgroup$ Commented Jan 2, 2018 at 3:25
  • $\begingroup$ Closely related: (9637), (38624), (157222). Possibly of interest: (19357) $\endgroup$ Commented Jan 3, 2018 at 16:54