Verification of answer in a birthday problem

Question

In the answer here, should the number of ways to pick the groups of triples, pairs and singlets from the 20 people be: $$\frac{20!}{\color{#C00}{3!^2}\,\color{#090}{2!^4}\,\color{#E90}{6!}}$$ since if you were to take this to the extreme and have 20 single people, there would be only 1 way to do it, which would be $$\frac{20!}{20!}$$

I would like to think it is a typo, but I want to make sure my understanding is correct.

Answer 1

The intended meaning of this quotient is elucidated in a comment under the answer by the author:

Consider the six people with unique birthdates. We choose the birthdates in chronological order, thus the $6!$ in the denominator. When we choose the $6$ people, we choose them in the $6!$ different orders, thus we only divide by $1!^6$ in the choice of people. In one or the other we need to divide by $6!$ or we end up choosing 6! orders of dates and 6! orders of people.

So if there were $20$ singletons, this quotient should indeed be $20!$ (and not $1$ as you suggest), since it accounts for the $20!$ different orders of the people. The denominator contains the factorial of each group size, since the order doesn’t matter within the groups, since all people in a group are assigned the same birthday.

As regards consistency with the other answer you linked to, you’d have to elaborate on where you see a contradiction.

Answer 2

The formula is not for "picking" twenty people, it is for the number of different partitions given the size of the elements of the partitions.

So it checks out that there is only one way to partition a set of twenty elements in twenty singletons.

Answer 3

I suspect you may have misread the linked answer.

If you want the $20$ people to all have different birthdays, the probability calculation is $\dfrac{20!}{1!^{20}}\,\dfrac{365!}{20!\, 345!} \, \dfrac1{365^{20}} \approx 0.58856$, somewhat more than $\frac12$ as you might expect from the birthday problem.

The full calculation for the probability that all $20$ share the same single birthday would be $\dfrac{20!}{20!^1}\,\dfrac{365!}{1!\, 364!} \, \dfrac1{365^{20}}$ which is $\frac{1}{365^{19}}\approx 2\times 10^{-49}$ and very small, again as you would expect.