Given 18 people, find the probability that among the twelve months, there are six containing two birthdays and six containing one. (You can assume each month has the same number of days)

The answer listed is here

I was wondering if my logic for each of the terms is correct. First I choose $6$ out of the $12$ months for one birthday. Then I choose $6$ out of the $18$ people to assign to those months, which can be arranged in $6!$ ways. Lastly with the remaining $12$ people I assign them to the two birthday months, with each month having two open slots, such that these people are arranged $12!$ ways. Then to account for the fact that the order of people in each of the $6$ months with two birthdays does not matter, I divide by $2!$ for each month, which equals $2!^6$. Ultimately, this is all over $12^{18}$, since that is the total ways to arrange the birthday. Does my logic seem correct to arrive to this answer? Would you have done it differently?

  • $\begingroup$ Yes, it all sounds fine. In the future, try to avoid linking to pictures and instead use MathJax and $\LaTeX$ to improve readability and avoid having to link to outside materials. $\endgroup$ – JMoravitz Jul 26 '17 at 15:26
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    $\begingroup$ The problem statement however could be improved, replacing the parenthetical phrase with a longer more accurate one such as "You can assume that the birthmonths for the 18 people are uniformly distributed and independently chosen." Simply assuming each month has the same number of days does not directly imply that they are equally likely to occur (more births occur during summer than during winter), also the mention of independence is necessary because otherwise we aren't told who these 18 people are. Maybe they were all taken from the May birthday list at their office, giving probability 0 $\endgroup$ – JMoravitz Jul 26 '17 at 15:32

Yes, that's correct. However, you can simplify one step slightly. There are $12^{18}$ ways of assigning months to people, and there are $\binom {12}6$ ways to choose which months have one person and which have two. Now once you've done that there are $18!$ ways to order the people (in terms of what order their birthdays come in the calendar), and each way of assigning people to months corresponds to $(2!)^6$ of these orderings, so the answer is $$\frac{18!\binom{12}{6}}{2^612^{18}}.$$ (This is the same as your terms, after applying the simplification $\binom{18}{6}6!12!=18!$.)


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