A) 46
B) 54
C) 60
D) None of these
Correct Answer: C
Solution :
[c] Let the sides of the game be A and B given 5 married couples, i.e, 5 husband and 5 wives. Now, 2 husbands for two sides A and B can be selected out of \[5{{=}^{5}}{{C}_{2}}=10\] ways. After choosing the two husbands their wives are to be excluded (since no husband and wife play in the same game). So we have to choose 2 wives out of remaining 5 - 2 = 3 wives i.e., \[^{3}{{C}_{2}}=3\] ways. Again two wives can interchange their sides A and B in 2! =2 ways. By the principle of multiplication, the required number of ways \[=10\times 3\times 2=60\]
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