5
$\begingroup$

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.

$\endgroup$
  • 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
4
$\begingroup$

Line up the girls.

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

Thus $Pr = \dfrac16$

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.