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Question

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?

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Solution

2 men are selected from 8 men in 8C2 ways since no husband and wife should be in same game.
2 woman out of remaining 6 are chosen in 6C2 ways.
Now one team can be chosen as (M1,W1) or (M1,W2) in 2 ways.
The required number of arrangements
=8C2×6C2×2
8!6!2!×6!4!2!×2
=8×7×6!6!×2!×6×5×4!4!×2!×2
4×7×3×5×2
= 840

1257537_1330261_ans_6221938f0daa48c197d95d5e5f3d2539.PNG

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