Distinguish nine-digit integers from nine-position binary strings

The printed answer 51 allows leading zeros. Place the four identical 1 s, creating five gaps for zero runs. No run may exceed length 2. Five isolated zero runs fill all five gaps in 1 way. One run 00 and three isolated zeros occupy four gaps in $\binom54\,4!/3!=20$ ways. Two runs 00 and one isolated zero occupy three gaps in $\binom53\,3!/2!=30$ ways. The printed total is $1+20+30=51$ nine-position binary strings.

There are $9!/(4!5!)=126$ unrestricted strings. Exclude the zero-run multiplicity patterns: [5] gives 5; [4,1] gives $\binom52\,2!=20$; [3,2] gives $\binom52\,2!=20$; and [3,1,1] gives $\binom53\,3!/2!=30$. Thus $126-5-20-20-30=51$. These are disjoint maximal-run patterns, not overlapping choices of a substring 000.

An ordinary nine-digit integer cannot start with 0. It must begin with 1, so only the four gaps after/between the four 1 s can contain zero runs. To place five zeros with each run at most 2, either the run sizes are a permutation of (2,1,1,1), giving 4 ways, or of (2,2,1,0), giving $4!/(2!1!1!)=12$ ways. Therefore the answer under the literal integer interpretation is $4+12=16$. The source’s 51 is correct only when these are nine-position strings with leading zeros allowed.

The eight circular objects are six distinct people and two indistinguishable vacancies. Fix one person as reference: $(8-1)!/2!=2520$.

Assign that reference person to one of eight numbered seats: $[(8-1)!/2!]\times8=20160$.