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Standard XII
Mathematics
nCr Definitions and Properties
In how many w...
Question
In how many ways can a mixed double game can be arranged from amongst
8
couples if no husband and wife play in the same game
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Solution
8
C
2
2 men are selected from 8 men
8
C
2
ways
Since no husband and wife should be in same game,
two women out of remaining 6 are chosen in
6
C
2
ways
Now one team can be chosen as
(
M
1
,
W
1
)
o
r
(
M
1
,
W
2
)
(M1,W1)or(M1,W2)
in 2 ways.
∴
∴
The required no. of arrangements =
8
C
2
*
6
C
2
*2
8C2×6C2×2
=
840
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