Please see example 22. I think second step is wrong. When 2 persons are specified,then remaining (n-2) should be arranged in (n-3)! ways. Isnt it?

Please see example 22. I think second step is wrong. When 2 persons are specified,then remaining (n-2) should be arranged in (n-3)! ways. Isnt it? n persons Can sit at a round table i_n (n — 1) ! ways
Number of exhaustive cases (n — 1) !
When two specified persons sit together, the remaining n — 2 persons along with the
two specified can sit in [In — 1) — 1! ! = (n — 2) ! ways.
The two specified persons can sit in 2 ! ways
. Number of favourable ways. i.e., number of ways when two specified persons sit
together = — 2) !
Probability that the two persons sit together
2
Odds against two specified persons sitting next to each Other are
n -1-2 '1-3
2
2