2012 AMC 12B Problem 12

Let $A$ be the sequences in which all zeros are consecutive and $B$ those in which all ones are consecutive. For $A,$ there is one all-ones sequence, $20$ sequences with exactly one zero, and $\binom{20}{2}=190$ sequences with two or more zeros (choose the first and last zero position). So $|A|=1+20+190=211,$ and by symmetry $|B|=211.$ A sequence in $A\cap B$ is a block of zeros followed by a block of ones, or the reverse; there are $|A\cap B|=40$ of these. Therefore $|A\cup B|=211+211-40$ $=382.$ Thus, the correct answer is E.