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

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