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In how many ways can a mixed doubles game in tennis can be arranged from 5 couples, if no husband and wife play in the same game?

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For a mixed doubles game 2 husbands and 2 wives are to be selected.
First of all 2 husbands can be selected in $^5C_2 $ ways.
Their respective wives are exempted in that game and
two wives are to be selected from the remaining 3 wives in $^3C_2$ ways.
Once the team is selected each team can be mixed in two ways, $i.e.,$ $H_1,W_1 $ and $H_1,W_2$
$\therefore\:$ The required ways $=^5C_2.^3C_2.2=60$.
answered Nov 23, 2013 by rvidyagovindarajan_1
edited Mar 12, 2014 by balaji.thirumalai

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