There are $2^12=4096$ little strings altogeher. That these, the variety of bit strings containing specifically 0,1 or 2 1"s is $$12choose 0+12choose 1+12choose 2=1+12+66=79$$Similarly, the variety of bit strings containing specifically 0,1 or 2 0"s is likewise 79. Over there is no overlap. Thenumber of little strings satisfying the problem is 4096 - 2 $ imes$79=4096-158=3938.

The means to ar only 2 "ones" (remaining gift "zeros") room $inom 122$, and $inom121, inom120$ those that placing only one and also none.Same for the "zeros".

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Taking out from the full $2^12= inom 120+ inom 121+ cdots$ the over two tails offers the answer.

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How plenty of bit strings of size 7 contain (a) exactly three 1s? (b) at many three 1’s? (c) at least three 1’s?

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