There are $6$ people, $3$ boys and $3$ girls. Each boy is in a relationship with one girl. Three pairs are randomly drawn. What is the probability that these three pairs will be the actual couples?

My reasoning was

$$ P = \frac{3}{\binom{6}{2}} \times \frac{2}{\binom{4}{2}} \times \frac{1}{\binom{2}{2}}$$

But this does not give me the answer a professor has given me. I'd appreciate some hints or new ways of approaching the problem.

  • 2
    $\begingroup$ When the first pair is “randomly drawn,” is it any of the $6\choose2$ pairs of two people, or is it any of the $3\times3$ pairs of a boy and a girl? You and the professor have interpreted this differently. $\endgroup$ – Steve Kass Mar 4 '16 at 14:21
  • $\begingroup$ The question explicitly states "Each boy is in a relationship with one girl" $\endgroup$ – true blue anil Mar 4 '16 at 17:19

Line up the girls.

There are $3! = 6$ ways of pairing the boys with the girls, only one of which is correct.

Thus $Pr = \dfrac16$

  • $\begingroup$ really nice answer $\endgroup$ – Bhaskara-III Mar 4 '16 at 14:04
  • $\begingroup$ @Bhaskara-III: Glad that you like it ! $\endgroup$ – true blue anil Mar 4 '16 at 14:19

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