At a dance there are three married couples: Ann and Andy, Betty and Bo

Question

At a dance there are three married couples: Ann and Andy, Betty and Boris, and Danielle and Dan. The wives select a husband at random with whom to share a dance. What is the probability that each of the three men dances with a woman other than his spouse.

A. 1/9
B. 2/9
C. 1/3
D. 1/2
E. 2/3

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Official Answer

C

A. 1/9
B. 2/9
C. 1/3
D. 1/2
E. 2/3

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TestPrepUnlimited · Answer

Although there are 6 people, there are not many possible outcomes since each pair must be husband + wife.

If we label the pairs A, B, C and start matchmaking for A, Wife A can dance with any of the 3 husbands. Wife B has only 2 more options afterwards and Wife C automatically takes the last husband. Thus there are only 6 options in total.

Among those, again starting from A, Wife A needs to choose either Husband B or C, and there is only one way for the rest of the people to fit the criteria after that selection for each selection. Thus 2 desired outcomes over 6 total, 1/3 chance.

Ans: C

BrentGMATPrepNow · Answer

Since there are very few outcomes in the scenario, a quick (and reliable) approach is to simply list all of the outcomes.
Let a, b, and c = the three men. 
Let A, B, and C = the three women.

The outcomes are: 
 1) a-A, b-B, and c-C (each spouse matched with their partner)
2) a-A, b-C, and c-B 
3) a-B, b-A, and c-C 
 4) a-B, b-C, and c-A 
 5) a-C, b-A, and c-B 
6) a-C, b-B, and c-A

So, there are 6 possible outcomes, and 2 of those outcomes are such that each of the three men dances with a woman other than his spouse.
So the probability = 2/6 = 1/3

Answer: C

GMATGuruNY · Answer

Let the proper ordering for the selection of husbands = ABD.

Number of ways to arrange the 3 letters A, B and D = 3! = 6
Of these 6 possible arrangements, only two have no husband in the correct position:
DAB
BDA

Thus:
P(no husband dances with his spouse) = 2/6 = 1/3

C

GMATGuruNY · Answer

Formula approach:

Let the correct ordering for the selection of husbands = ABD.
Number of ways to arrange the 3 husbands = 3! = 6

A DERANGEMENT is a permutation in which NO element is in the correct position.
Given n elements (where n>1):
Number of possible derangements =  n! * (\frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} +  ... +  \frac{(-1)^n}{n!})

Thus, given 3 letters A, B and D:
Number of possible derangements =  3! * (\frac{1}{2!} - \frac{1}{3!}) = 3-1 = 2

Since there are 6 possible arrangements for the 3 husbands A, B and D -- and 2 possible derangements -- P(no husband is in the correct position) = 2/6 = 1/3

C

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At a dance there are three married couples: Ann and Andy, Betty and Bo

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At a dance there are three married couples: Ann and Andy, Betty and Bo

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