In how many ways can a lawn tennis mixed double be made up from seven married couples if no husband and wife play in the same set?

We arrange any 2 men in ${ }^{\prime} P_{2}$ ways and then the wives of the remaining 5 men can be arranged in ${ }^{5} P_{2}$ ways. This is because these two men should not be with their respective wives.

$\therefore$ By fundamental principle of counting, the required number of ways $={ }^{7} P_{2} \times{ }^{5} P_{2}$

$=\frac{7 !}{5 !} \times \frac{5 !}{3 !}$

$=\frac{7 !}{3 !}$

= 840