There space $2^12=4096$ little strings altogeher. That these, the number of bit strings containing specifically 0,1 or 2 1"s is $$12\choose 0+12\choose 1+12\choose 2=1+12+66=79$$Similarly, the number of bit strings containing specifically 0,1 or 2 0"s is also 79. There is no overlap. Thenumber of bit strings to solve the condition is 4096 - 2 $\times$79=4096-158=3938.
The means to ar only two "ones" (remaining being "zeros") are $\binom 122$, and also $\binom121, \binom120$ those that placing only one and none.Same for the "zeros".
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Taking out from the complete $2^12= \binom 120+ \binom 121+ \cdots$ the above two tails offers the answer.
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