Q:In how many ways , we can choose two teams of mixed double for a tennis tournament from four couples such that if any couple participates , then it is in the same team . NOTE:-[Please do not give me the link in which this question is answered wrongly with answer 6 , however the answer is 42] Share with your friends Share 8 Ashish Shukla answered this Dear Student, First select the 2 men this can be done in 4C2 ways. Let's denote the four couples by (A,a), (B,b), (C,c), and (D,d) where A,B,C,D are the husbands and a,b,c,d are the wives. For a games, let's choose first the husbands. We have 4C2=4×32=6 possibilities. Now we know that the final number of games will be a multiple of 6, . Once husbands were chosen, say A and B, let's count how many possibilities we have to choose their partners under the given restrictions. We can choose a and b and we have to pair them as (A,a) and (B,b) - 1 possibility. We can choose one of the wives, a or b, but we have to pair her with the her husband and in addition, we have to choose another partner for the second husband. This we can do in 2*2 = 4 ways, as there are two possibilities to choose from a and b, then we have 2 possibilities to choose the other wife, c or d (A,c) and (B,b), (A,d) and (B,b), (A,a) and (B,c), (A,a) and (B,d)- 4 possibilities , we can choose the other two wives, c and d, and we have two possibilities to team them up with the men, (A,c), (B,d) or (A,d), (B,c) - 2 possibilities. In conclusion, for every pair of husbands, we have 1 + 4 + 2 = 7 possibilities to choose their partners for the game. Total number of possibilities 6 * 7 = 42.Regards. -12 View Full Answer
