For a mixed doubles tennis game (one gent and one lady on each side) 4 persons are selected at random from ten married couples. Then the probability that the selection doesn’t contain a couple (a husband and his wife) is

n(s)= No. of ways of choosing 2 men, 2 women, and
dividing them into two teams $=\left({ }^{10}{\mathrm{C}}_2{\times }^{10}{\mathrm{C}}_2\times 2\right)$
No. of favourable cases
= No. of ways of choosing 2 men, excluding their
wives choosing 2 women, and dividing into two
teams $=\left({ }^{10}{\mathrm{C}}_2{\times }^8{\mathrm{C}}_2\times 2\right)$
$\therefore$Required probility $=\frac{{ }^8{\mathrm{C}}_2}{{ }^{10}{\mathrm{C}}_2}=\frac{28}{45}$