(a) 840
We can choose 2 men out of 8 men in 8C2 ways. Since no husband and wife are to play in the same game, two women out of the remaining 6 women can be chosen in 6C2 ways. If M1, M2, W1, W2 are respectively the two men and two women chosen for the teams then the teams can be formed as (M1, W1) (M2, W2) or (M1, W2) (M2, W1), i.e., in 2 ways.
Hence, total number of ways of arranging the game = 8C2 × 6C2 × 2
= \(\frac{8\times7}{2}\times\frac{6\times5}{2}\) = 840.