For a positive integer
define a sequence of zeros and ones to be balanced if it contains
zeros and
ones. Two balanced sequences
and
are neighbors if you can move one of the
symbols of
to another position to form
. For instance, when
, the balanced sequences
and
are neighbors because the third (or fourth) zero in the first sequence can be moved to the first or second position to form the second sequence. Prove that there is a set
of at most
balanced sequences such that every balanced sequence is equal to or is a neighbor of at least one sequence in
.
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For a positive integer $n$ define a sequence of zeros and ones to be balanced if it contains $n$ zeros and $n$ ones. Two balanced sequences $a$ and $b$ are neighbors if you can move one of the $2n$ symbols of $a$ to another position to form $b$. For instance, when $n = 4$, the balanced sequences $01101001$ and $00110101$ are neighbors because the third (or fourth) zero in the first sequence can be moved to the first or second position to form the second sequence. Prove that there is a set $S$ of at most $\frac{1}{n+1} \binom{2n}{n}$ balanced sequences such that every balanced sequence is equal to or is a neighbor of at least one sequence in $S$.