One Indian and four American men and their wives are to be seated...

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 $$\begin{aligned}n(E) & =(5 !) \times(2 !)^4 \\ P\left(\frac{A}{E}\right)& =\frac{n(A \cap E)}{n(E)}\\ & =\frac{(4 !) \times(2 !)^5}{(5 !) \times(2 !)^4}=\frac{2}{5} \end{aligned}$$\Rightarrow \quad$P\left(\frac{A}{E}\right)=\frac{n(A \cap E)}{n(E)}$ALITER 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 favourable. $\therefore$ Required probability $=\frac{2}{5}$.