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I am stuck, and would like someone to lead me in the right direction.

Given thirty people, find the probability that among twelve months there are six containing two birthdays and six containing three.

Solution.

I would compute the probability of this event as:

$$p=\frac{{30 \choose 2}{12 \choose 6}6^{2}{28 \choose 3}{6 \choose 6}6^{3}}{12^{30}}$$

Am I counting correctly? My answer doesn't match that given in the text.

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    Your denominator $12^{30}$ looks fine, but please explain how you came up with that numerator. I get $$\binom{12}6\binom{30}2\binom{28}2\binom{26}2\binom{24}2\binom{22}2\binom{20}2\binom66\binom{18}3\binom{15}3\binom{12}3\binom93\binom63\binom33$$ $$=\frac{12!}{6!6!}\cdot\frac{30!}{(2!)^6(3!)^6}.$$ – 2017-01-17
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    Why don't you tell us the answer in the text, so we don't have to look it up? – 2017-01-17
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    @bof, the answer in the text is precisely what you have written, $$\frac{30!}{{2^6}{6^6}}{12\choose6}12^{-30}$$ I am curious to know, why do we keep sampling sequentially 2 people from the group of 30 people until we hit 18, and then sample sequentially in 3's? – 2017-01-17
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    First, in $\binom{12}6$ ways, we decide which months have two birthdays. Say we decide that the first six months of the year have two birthdays. Then, from the $30$ people, we have to decide which two have January birthdays; next, from the $28$ remaining people, which two have February birthdays, and so on. – 2017-01-17
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    @bof, I see the logic now. So, that's how $2(6)+3(6)=30$ birthdays in all. I should have interpreted the problem as - among 12 months there are 6 months that contain 2 birthdays "each" and 6 months containing 3 birthdays "each". I counted for only one of $2$ and $3$ birthday months. – 2017-01-17
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    @Quasar - Using stars and bars, number of ways for birthday-month combination is $\binom {30+11}{11}=\binom {42}{11}$. Number of ways of having $6$ months with $2$ birthdays and $6$ months with $3$ birthdays is $\binom {12}6$. Hence probability required is $\binom {12}6\big/\binom {42}{11}$. Why does this not work? – 2017-01-17
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    @hypergeometric, if we had $r=30$ indistinguishable people(balls) to be placed in $n=12$ cells, the number of distinguishable distributions are - $${{r+n−1}\choose{n−1}}$$ Any distribution can be expressed in the form of occupancy numbers of the cells $1,2,…,12$. The above expression for $\text{# of ways to place r indistinguishable balls in n cells}$ would also include distributions such as $(30,0,0,0,0,0,0,0,0,0,0,0)$. These do not satisfy the criteria given in the problem. – 2017-01-17
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    @hypergeomtric, the distinguishable outcomes, taking the people to be indistinguishable, are not all equally likely. "30 in January" is a lot less likely than "2 each Jan-Sept. and, 4 each Oct-Dec." It's these unequally likely outcomes that are counted by stars and bars (given the 30 birthdays are each uniformly and independently distributed among the 12 months). – 2018-11-12

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