Birthday problem-Probability exactly $2$ triples and $4$ pairs if $20$ people in room

Question

Say there are 20 people in a room. What is the probability there are exactly 2 triples and 4 pairs. Is my answer shown below correct? Assume 365 days in the year.

$P= \dfrac{\binom{365}{2}\binom{363}{4}\binom{20}{3}\binom{17}{3}\binom{14}{2}\binom{12}{2}\binom{10}{2}\binom{8}{2} \cdot 359 \cdot 358 \cdot 357 \cdot 356 \cdot 355 \cdot 354}{365^{20}}$?

Term $365C2$ chooses the $2$ birthdays for the $2$ triples. Each triple has a different birthday. Term $363C4$ chooses the $4$ birthdays for the $4$ pairs. Each pair has different birthdays. Term $20C3$ selects the $3$ people for the first triple and $17C3$ the $3$ people for the second triple. Term $14C2$ picks the $2$ people for first pair, $12C2$ for second pair, $10C2$ the third pair, and finally $8C2$ for the fourth pair. The term $(359 \cdot 358 \cdot 357 \cdot 356 \cdot 355 \cdot 354)$ is the birthdays for the remaining $6$ people, which do not match. I start with $359$ because $6$ birthdays have taken by the $2$ triples and the $4$ pairs. All this is then divided by the total number of possible birthday selections $365^{20}$.

I am wondering if the selection of the people for the $2$ triples should be $20C6$ instead of $20C3 \cdot 17C3$ as I show. I believe my method is the correct one. Please let me know.

Answer 1

You have solved the problem correctly.

There are $\binom{365}{2}$ ways to choose the two days for the triples. There are $\binom{20}{3}$ ways to choose which three people share the earlier of these two birthdays and $\binom{17}{3}$ ways to choose which three of the remaining people share the later of these two birthdays.

If you chose which six people were part of the triples, you would have to multiply the $\binom{20}{6}$ ways of selecting the six people by the $\binom{6}{3}$ ways of selecting which three of those six people had the earlier birthday. Notice that $$\binom{20}{3}\binom{17}{3} = \frac{20!}{3!17!} \cdot \frac{17!}{3!14!} = \frac{20!}{3!3!14!} = \frac{20!}{6!14!} \cdot \frac{6!}{3!3!} = \binom{20}{6}\binom{6}{3}$$

Answer 2

The probability of getting $\color{#C00}{2\text{ triple}}$ birthdays, $\color{#090}{4\text{ double}}$ birthdays, and $\color{#E90}{6\text{ single}}$ birthdays out of $20$ people: $$ \overbrace{\quad\frac{20!}{\color{#C00}{3!^2}\,\color{#090}{2!^4}\,\color{#E90}{1!^6}}\quad}^{\substack{\text{number of ways to pick}\\\text{the groups from the $20$}\\\text{people}}}\overbrace{\frac{365!}{\color{#C00}{2!}\,\color{#090}{4!}\,\color{#E90}{6!}\,353!}}^{\substack{\text{number of ways to}\\\text{pick the dates for}\\\text{each group}}}\overbrace{\ \ \frac1{365^{20}}\ \ }^{\substack{\text{number of}\\\text{ways to pick}\\\text{$20$ birthdays}}}\tag1 $$ which equals $$ \frac{499036100988931803494442957177962496}{1544238596637480626819418327224222027587890625}\tag2 $$ or approximately $3.2315997157146795\times10^{-10}$.

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Explanation Of The Multinomial Coefficients

For given permutations of the $365$ dates in a year and of the $20$ people, here is a choice of the $\color{#C00}{2\text{ triple}}$ birthdates, the $\color{#090}{4\text{ double}}$ birthdates, and the $\color{#E90}{6\text{ single}}$ birthdates, and the people who have those birthdates: $$\newcommand{\date}[3]{\scriptsize\color{#1}{\!\!#2\text{ #3}\!\!}} \begin{array}{|c|c|} \hline \,\date{#C00}{2}{Jan}&\date{#C00}{20}{Mar}&\date{#090}{15}{Feb}&\date{#090}{7}{Apr}&\date{#090}{30}{May}&\date{#090}{11}{Sep} &\date{#E90}{22}{Jan}&\date{#E90}{28}{Feb}&\date{#E90}{24}{Jun}&\date{#E90}{13}{Oct}&\date{#E90}{17}{Nov}&\date{#E90}{25}{Dec}&\scriptsize\!\!353\text{ left}\!\\ \hline 6&2&7&9&11&10&4&16&3&14&15&1\\ 12&5&8&20&17&18\\ 19&13\\ \hline \end{array} $$ Note that the dates in each color are in chronological order; thus, the number of such choices is $$ \frac{365!}{\color{#C00}{2!}\,\color{#090}{4!}\,\color{#E90}{6!}\,353!}\tag3 $$ Furthermore, the numbers in each column are in numerical order; thus, the number of such fillings of the columns is $$ \frac{20!}{\color{#C00}{3!^2}\,\color{#090}{2!^4}\,\color{#E90}{1!^6}}\tag4 $$

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The Results Match

Note that we can write the multinomial coefficients as products of binomial coefficients: $$ \frac{20!}{3!^2\,2!^4\,1!^6} =\textstyle\binom{20}{3}\binom{17}{3}\binom{14}{2}\binom{12}{2}\binom{10}{2}\binom{8}{2}\color{#C00}{\overbrace{\binom{6}{1}\binom{5}{1}\binom{4}{1}\binom{3}{1}\binom{2}{1}\binom{1}{1}}^{6!}}\tag5 $$ and $$ \frac{365!}{2!\,4!\,6!\,353!}=\textstyle\binom{365}{2}\binom{363}{4}\color{#C00}{\binom{359}{6}}\color{#CCC}{\binom{353}{353}}\tag6 $$ The product of the red terms is $6!\binom{359}{6}=359\cdot358\cdot357\cdot356\cdot355\cdot354$ and the greyed out term is $1$.

Thus, the result in $(1)$ can be written as $$ \frac{\binom{365}{2}\binom{363}{4}\binom{20}{3}\binom{17}{3}\binom{14}{2}\binom{12}{2}\binom{10}{2}\binom{8}{2}\cdot359\cdot358\cdot357\cdot356\cdot355\cdot354}{365^{20}}\tag7 $$ which matches the result in the question.