In how many ways can a mixed doubles tennis g | vedclass

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(D) To form a mixed doubles game,we need $2$ men and $2$ women such that no husband and wife are paired together.First,select $2$ men out of $7$ in ${

Vedclass · Accepted Answer

$420$

Vedclass · Answer

(D) To form a mixed doubles game,we need $2$ men and $2$ women such that no husband and wife are paired together.First,select $2$ men out of $7$ in ${ }^7 C_2$ ways.Number of ways to select $2$ men $= \frac{7 	imes 6}{2 	imes 1} = 21$.Now,we have $7$ women available. Since the $2$ men selected have their wives in the group,we must exclude these $2$ wives to ensure no husband and wife are in the same game.Remaining women available $= 7 - 2 = 5$.Select $2$ women out of $5$ in ${ }^5 C_2$ ways.Number of ways to select $2$ women $= \frac{5 	imes 4}{2 	imes 1} = 10$.Now,we have $2$ men $(M_1, M_2)$ and $2$ women $(W_1, W_2)$.We need to form $2$ mixed doubles pairs. The first man $M_1$ can be paired with either $W_1$ or $W_2$ ($2$ ways),and the second man $M_2$ will be paired with the remaining woman ($1$ way).Total ways $= { }^7 C_2 	imes { }^5 C_2 	imes 2 = 21 	imes 10 	imes 2 = 420$.

In how many ways can a mixed doubles tennis game be arranged from $7$ married couples if no husband and wife are in the same game?

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