• # question_answer The number of ways in which a mixed double game can be arranged from 9 married couples if no husband and wife play in the same game is A) 756                               B) 1512 C) 3024                             D) None of these

[b] We have, 9 married couples. We can select two men out of 9 in ${}^{9}{{C}_{2}}$ ways. Since no husband and wife are to play in the same game, two women out of the remaining 7 can be chosen in ${}^{7}{{C}_{2}}$ways. If ${{M}_{1}},$${{M}_{2}},$${{W}_{1}}$${{W}_{2}}$ are chosen, then a team consist of ${{M}_{1}}{{W}_{1}}$ and ${{M}_{2}}{{W}_{2}}.$ Thus, the number of ways of arrangement is ${}^{9}{{C}_{2}}\times {}^{7}{{C}_{2}}\times 2=1512.$