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I am trying to figure out how many 13-card bridge hands, drawn from an ordinary 52-card deck, contain one, two, or three quads, where a "quad" is all four cards of the same rank. This is not pertinent to bridge, but it's very pertinent to cribbage, in which one one fourth of the deck is ever dealt out in any given hand (i.e., six cards to each player and a starter card). Intuitively, all four of a given rank should show up fairly rarely when only a fourth of the deck is dealt out, and having two or three sets of quads should be very rare indeed.

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    $\begingroup$ If $k$ is the number of quads in a hand, there are $13 \choose k$ sets of $k$ quads or in other words, $13 \choose k$ unordered hands of size $4k$ with $k$ quads. Now for each $k$, just find the number of hands (after removing $4k$ cards, your quads, from the deck) of size $13-4k$. For instance in $k=3$ case, you should get ${13 \choose 3}40$. $\endgroup$ – PVAL-inactive Sep 18 '17 at 22:54
  • $\begingroup$ This is a job for inclusion-exclusion $\endgroup$ – Thomas Andrews Sep 18 '17 at 22:54
  • $\begingroup$ @PVAL-inactive: if $k = 4$, there cannot be $k$ quads in a hand, but $C_4^{13} \neq 0$. $\endgroup$ – Rob Arthan Sep 18 '17 at 23:00
  • $\begingroup$ @RobArthan That is taken into account by "of size $13-4k.$" When $k>3$ then $13-4k<0$ and you can't get any such cards. (But PVAL's hint has other problems of over-counting, so you need inclusion-exclusion.) $\endgroup$ – Thomas Andrews Sep 18 '17 at 23:09
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Let $A_i$ be the set of hands that have a quad of rank $i=1,\dots,13$. Then inclusion-exclusion has:

$$\begin{align}\left|A_1\cup \cdots \cup A_{13}\right|&=\sum_{i=1}^{13}|A_i|-\sum_{1\leq i<j\leq 13} |A_i\cap A_j| + \sum_{1\leq i<j<k\leq 13} |A_i\cap A_j\cap A_k|\\ &=\binom{13}{1}\binom{48}{9}-\binom{13}{2}\binom{44}{5}+\binom{13}{3}\binom{40}{1} \end{align}$$

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