2012 AMC 12B 第 12 题

Let $A$ be the sequences with all zeros together and $B$ those with all ones together, then use $|A\cup B|=|A|+|B|-|A\cap B|$

A sequence with the zeros together is fixed by choosing the first and last zero positions, giving $1+20+\binom{20}{2}$ of them

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.