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In how many ways a mixed double games can be arranged from 8 married couples if no husband and wife play in the same game?

$(A)\; 420 \\(B) 210 \\(C) 840 \\(D) 105 $

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2 men are selected from 8 men in $^8C_2$ ways.
Since no husband and wife should be in same game,
two women out of remaining 6 are chosen in $^6C_2$ ways.
Now one team can be chosen as $(M_1,W_1)\:or\:(M_1,W_2)$ in 2 ways.
$\therefore$ The required no. of arrangements = $^8C_2\times ^6C_2\times 2$
$=840 $
answered Sep 5, 2013 by rvidyagovindarajan_1
 

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