Explanation

(c)
Let $E =$event when each American man is seated adjacent to his wife
and $A =$event when Indian man is seated adjacent to his wife.
Now, $n(A \cap E) = (4!) \times (2!)^{5}$
Even when each American man is seated adjacent to his wife.
Again, $n(E) = (5!) \times (2!)^{4}$
$\therefore P\left( \frac{A}{E} \right) = \frac{n(A \cap E)}{n(E)}$
$= \frac{(4!) \times (2!)^{3}}{(5!) \times (2!)^{4}} = \frac{2}{5}$