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Bunuel
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At a dance there are three married couples: Ann and Andy, Betty and Bo 17 Jan 2021, 09:12
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57% (01:43) correct 43% (01:50) wrong based on 220 sessions

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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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At a dance there are three married couples: Ann and Andy, Betty and Bo 17 Jan 2021, 09:34
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Bunuel
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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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
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At a dance there are three married couples: Ann and Andy, Betty and Bo 13 Feb 2022, 14:55
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Bunuel
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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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
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At a dance there are three married couples: Ann and Andy, Betty and Bo Updated on: 13 Feb 2022, 16:08
Originally posted by GMATGuruNY on 13 Feb 2022, 15:52.
Last edited by GMATGuruNY on 13 Feb 2022, 16:08, edited 1 time in total.
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Bunuel
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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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

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At a dance there are three married couples: Ann and Andy, Betty and Bo 13 Feb 2022, 16:07
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Bunuel
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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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

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At a dance there are three married couples: Ann and Andy, Betty and Bo 25 Oct 2025, 13:50
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