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Relatively simple-sounding question:

I know the Birthday Paradox, how it works, and how you calculate the probability of at least or exactly $n$ people sharing the same birthday in a group of $m$ people.

But now I wonder, how do you calculate the probability of $n$ pairs of people who share the same birthday, but may have different ones to other pairs, within a group of $m$ people?

And then of course the extension of that with not just pairs but a group of $i$ people. Which I just realized would be exactly the standard Birthday Paradox if you let $i=1$.

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    $\begingroup$ Hi, are you asking for $n$ disjoint pairs? Or can the pairs overlap, in which case do you count a triplet as 3 pairs and a quartet as 7? $\endgroup$
    – Robertmg
    Commented Sep 5 at 3:31
  • $\begingroup$ @Robertmg Wouldn't a quartet be $\color{red}{6}$? $\endgroup$
    – K. Jiang
    Commented Sep 5 at 4:56
  • $\begingroup$ @K.Jiang yes sorry small typo. Thank you. $\endgroup$
    – Robertmg
    Commented Sep 5 at 12:42
  • $\begingroup$ oeis.org/A014088 $\endgroup$
    – awkward
    Commented Sep 5 at 13:14

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