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HamedATX

I think the answer is A=1/8.
The total number of ways 4 women can dance with the 4 men is 4!=24 and there is only 3 ways that none of the 4 wives won't dance with her souse. So the answer is 3/24 = 1/8

There are 3 options and 3 ways that none of the 4 wives won't dance with her souse. So the answer is 3*3/24 = 9/24.
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I thought, the first wife can choose from 3 men other than her husband - so that probability would be 3/4
Similarly, second wife can choose from 2 other men from remaining 3 men - that probability would be 2/3
After that, the third wife can choose one from the remaining 2 - that probability would be 1/2

So the each of the four men dances with a woman other than his spouse - 3/4*2/3*1/2*1/1 = 1/4 - where am I wrong?
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Stalin1994
I thought, the first wife can choose from 3 men other than her husband - so that probability would be 3/4
Similarly, second wife can choose from 2 other men from remaining 3 men - that probability would be 2/3
After that, the third wife can choose one from the remaining 2 - that probability would be 1/2

So the each of the four men dances with a woman other than his spouse - 3/4*2/3*1/2*1/1 = 1/4 - where am I wrong?
­The above may not be correct always..

say A1 chooses one from B2, C2 and D2.
If it is B2, then B1 also has three to choose from A2, C2 and D2.
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No matches =

Total matches possible - (All match + 2 match + 1 match)

Total matches = 4! = 24

All match = 1 way

Two matches:

Number of ways to pick a matching pair = 4!/2!2! =

6 ways

One match:

ABCD
BCAD
CABD

Using D as the match, 2 ways.

How many ways to pick the one to match: 4

Total ways: 4*2 = 8


Total ways for no matches:

24-1-6-8 = 9

Probability: 9/24

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P=D(4)/4!=9/24­

D(n) is dearrangements
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