The number of ways in which a mixed double game can be arranged from amongst 5 married couples, if no husband and wife play in the same game is:

Let sides of game A and B, given 5 married couples, i.e. 5 husbands’ and 5 wives. Now 2 husband for two sides A and B be selected out of 5 = C (5, 2) = 10ways after choosing the two their wives are to be excluded (since no husband and wife play in the same game). So we are to choose 2 wives out of remaining 5−2=3 wives, i.e. C(3,2)=3 ways.Again two wives can interchange their sides A and B is = 2waysSo required no. of ways = 10×3×2=60