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I have been doing a bunch of birthday problem questions, however this one has thrown a mental block my way. The questions is:

What is the probability that in a room of 10 people exactly 3 people will have the same birth month as each other, while the other 7 have different birth months from everyone else.

Tom
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2 Answers2

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The number of ways to choose $3$ out of $10$ people is $\binom{10}{3}$.

The number of ways to choose $8$ out of $12$ birth-months is $\binom{12}{8}$.

The number of ways to assign $8$ birth-months among them is $8!$.

The total number of ways for $10$ people to have birth-months is $12^{10}$.

So the probability is:

$$\frac{\binom{10}{3}\cdot\binom{12}{8}\cdot8!}{12^{10}}\approx3.868\%$$

Please note that by choosing $8$ birth-months, we are essentially choosing $1$ birth-month for the group of $3$ people, and $7$ different birth-months for each one of the remaining $7$ people.

barak manos
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the answer is $0.03868$ which is same as posted by Barak Manos.

firstly the problem assumes that all people in the room are born in the same year. $10$ people can choose $12$ months in $12^{10}$ ways. This forms our sample space.---(1)

Now as the problem states that exactly $3$ people have their b'days in the same month, first we choose $3$ people from the $10$ people in $10C3$ ways. now as the $3$ people have their b'days in a particular month, that month can be selected in $12C1$ ways. for the remaining $7$ people there are $11$ months left. And since their b'days must not fall in the same month, the choosing of months is without replacement.

So the numerator becomes: $10C3 * 12C1 *(11C1 * 10C1 * 9C1 * 8C1 * 7C1 * 6C1 *5C1)$....(2)

Note that the term in the bracket indicates the remaining $7$ people choosing their b'days.

So the required probability is (2) divided by (1)

N. F. Taussig
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Soumojit Bose
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    A well described analysis. I think the $3!$ term should be removed. – André Nicolas Mar 31 '15 at 16:19
  • The idea is that we line up the people in some arbitrary way (say by weight) and ask their birth month. We get $12^{10}$ possible sequences of length $10$ made up of months. Now we count the "good" sequences. Coose the popular month, then choose the $3$ places where it occurs, then fill in the rest with distinct months. No $3!$ term. – André Nicolas Mar 31 '15 at 16:26
  • If you do the calculation without the $3!$, you will get about $0.03868$. – André Nicolas Mar 31 '15 at 16:38
  • The only thing you added to my explanation is "taking these $3$ people in a group, these can arrange themselves in $3!$ ways", which is the wrong this to do in the context of the question at hand! – barak manos Mar 31 '15 at 16:52
  • okay....i do not want to start a bickering match here. But I must ask where did you get the 8! term from?? – Soumojit Bose Mar 31 '15 at 17:20
  • You've got $8$ groups: $1$ group of $3$ people, and $7$ groups of $1$ person each. You need to choose $8$ out of $12$ months, and then permute these $8$ months. – barak manos Mar 31 '15 at 17:51
  • @barak: I wrote that 3! by mistake. apart from that if you see the way that I approached the problem( it's explained), I don't think you'll find any queries. the (111098765) shows that each of the 7 members(left after the 3 members with the same month birthday) have 11 then 10 then 9 then 8....and so on options out of which to choose a month(because each of these 7 members have birthdays in 7 different months). – Soumojit Bose Mar 31 '15 at 17:56
  • @AndréNicolas: 3! duly removed. :-) – Soumojit Bose Mar 31 '15 at 18:01
  • @barakmanos: buddy..I meant no disrespect. I just cudn't figure out before that you got the 8! term. Now it has been explained. :-) P.s.- Do check out the previous comment. And do delete your last comment. – Soumojit Bose Mar 31 '15 at 18:14
  • @SoumojitBose: And the answer duly upvoted. – André Nicolas Mar 31 '15 at 22:08
  • Please see this tutorial on how to typeset mathematics on this site. – N. F. Taussig Apr 01 '15 at 10:33