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Re: At a dance there are four married couples: Ann and Andy, Betty and Bor [#permalink]
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HamedATX wrote:
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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Re: At a dance there are four married couples: Ann and Andy, Betty and Bor [#permalink]
chetan2u wrote:
Bunuel wrote:
At a dance there are four married couples: Ann and Andy, Betty and Boris, Danielle and Dan, and Elaine and Eric. The wives select a husband at random with whom to share a dance. What is the probability that each of the four men dances with a woman other than his spouse?

A. 1/8
B. 1/6
C. 1/4
D. 1/3
E. 9/24



This is a derangement question, that is questions where none element is in its original position.
Direct formula for n elements = \(n!(\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}....+\frac{1}{n!})\)
Here n = 4 => \(4!(\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!})\)
\((\frac{4!}{2!}-\frac{4!}{3!}+\frac{4!}{4!})=12-4+1=9\)

Total ways .. First lady can choose in 4 ways, the next in 3 ways and so on => 4*3*2*1=24

P = 9/24

E




Can you ellobarate the formula?

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Re: At a dance there are four married couples: Ann and Andy, Betty and Bor [#permalink]
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Re: At a dance there are four married couples: Ann and Andy, Betty and Bor [#permalink]
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