0
$\begingroup$

In a room of 30 people what is the probability that there are only 3 groups of 3 people where everyone in each group shares the same birthday and where the birthday shared in one group is different to the other two? (No more than three groups no less) Assumptions: 1) The remaining 21 people can also share birthdays. 2) They can also share birthdays in groups larger than 2.

$\endgroup$
  • $\begingroup$ Welcome to MathSE. What have you tried? $\endgroup$ – jvdhooft Sep 5 '19 at 14:16
  • $\begingroup$ I am not a mathematician but I have tried a variation of it. I don't know whether it's right, $\endgroup$ – Deepthinker101 Sep 5 '19 at 14:24
0
$\begingroup$

Assuming that the remaining $21$ people must all have different birthdays:

  • Pick the three repeated birthdays simultaneously.

  • Pick the three people who all share the earliest of those selected birthdays.

  • Pick the three people who all share the second of the selected birthdays.

  • Pick the three people who all share the final of the selected birthdays.

  • From youngest to oldest of the remaining people, choose their birthday, keeping in mind that it may not be any previously selected birthday.

Applying multiplication principle, we then find the number of different distributions of birthdays. Assuming each of the available birthdays are equally likely to occur and birthdays are independently chosen for each person, then the $365^{30}$ arrangements are equally likely and so we divide to get our final result.

$$\dfrac{\binom{365}{3}\binom{30}{3}\binom{27}{3}\binom{24}{3}362\frac{21}{~}}{365^{30}}$$ where $n\frac{r}{~}$ represents the falling factorial

Allowing the remaining 21 people to potentially share a birthday with another, begin the same way, but after having decided on the birthdays and people for your groups of three, break into cases based on the number of birthdays shared by two people.

With $k$ birthdays being shared by two people:

  • Pick which $k$ birthdays those are from those remaining available dates

  • Pick which two remaining people have the earliest of the selected birthdays

  • Pick which two remaining people have the next earliest of the selected birthdays

  • $\vdots$

  • Finally, pick which birthday each remaining person has from youngest to oldest.

$$\dfrac{\binom{365}{3}\binom{30}{3}\binom{27}{3}\binom{24}{3}\left(\sum\limits_{k=0}^{10}\left(\prod\limits_{i=1}^k\binom{23-2i}{2}\right)(362-k)\frac{(21-2k)}{~}\right)}{365^{30}}$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Hi can you check wether my answer to a variation I did is correct? $\endgroup$ – Deepthinker101 Sep 5 '19 at 14:26
  • $\begingroup$ Also the assumption is not part of my question sorry to not have made that clear. $\endgroup$ – Deepthinker101 Sep 5 '19 at 14:29
  • $\begingroup$ They can share birthdays too $\endgroup$ – Deepthinker101 Sep 5 '19 at 14:30
  • $\begingroup$ In that case it gets incredibly tedious as the simplest approach I can think of off the top of my head is to break into cases based on the number of people who share a birthday with one other person $\endgroup$ – JMoravitz Sep 5 '19 at 14:37
  • $\begingroup$ Tedious... I thought all you mathematicians were IQ160+ $\endgroup$ – Deepthinker101 Sep 5 '19 at 14:46

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.