Let E = event when each American man is seated adjacent to his wife..
A = event when Indian man is seated adjacent to his wife
Now, n(A ∩ E) = (4!)×(2!)^5
Even when each American man is seated adjacent to his wife Again n(E) =(5!)×(2!)^4
P(A/E)=(n(A∩E))/n(E)
=(4!)×(2!)^5/(5!)×(2!)^4
=2/5.
Alternative Fixing four American couples and one Indian man in between any two couples; we have 5 different ways in which his wife can be seated, of which 2 cases are favorable.
∴ required probability = 2/5