$(a)\;[8,9]\qquad(b)\;[10,12)\qquad(c)\;(11,13]\qquad(d)\;(14,17)$

Since each men play with each other the combination will be x^C_2 , but there are two games for each then 2* x^C_2

Each man plays with the two women, number of ways would be 2x.

Given 2 * x^C_2 -2x =66

x^2-3x-66=0

integer x should lie between (b) to exceed games by 66.

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