Number of 7-card hands with at least one exact pair? I checked the related questions How many ways to get at least one pair in a seven card hand? and Dealing a 5 card hand with exactly 1 pair, but didn't see an answer to my precise question. 
It's easy to count the number of 7-card hands from a 52-card pack with at least one pair or triplet or 4-of-a-kind (quadruplet?): $C(52,7)-4^7C(13,7)$, with the latter term being the number of hands that have no repeated ranks. The fraction of hands with at least one such multiplet is $508027\,/\,643195$ or about $0.790$.
I want the number of hands that have at least one pair that is not also a triplet and not also a quadruplet. Empirically, the fraction of such hands is about $0.738$. 
My unsuccessful attempt at a theoretical calculation goes as follows:
There are $52$ choices for the first card. To complete the mandatory pair, take one of the three remaining cards of the same rank and different suits (three choices), and eliminate the remaining two of the same rank, leaving $48$ cards. Draw the remaining five cards in $48!\,/\,43!$ different ways (multiplets in the remaining five are fine). Reckon as equivalent all $7!$ permutations of the total hand, for a final count of 
$({52\times{3}\times{48}\times{47}\times{46}\times{45}\times{44}})\,/\,{7!}$
Unfortunately, this is $44519904\,/\,7$, not only non-integer, but yielding a fraction $0.0475$ of $C(52,7)$, much too low.
What is wrong with my counting argument?
What is the right counting argument?
 A: I'd prefer to count directly for each "good" pattern, by multiplying two factors, 
[Choose various ranks]$\times$ [Choose suits for chosen ranks]
$4-2-1: [\binom{13}{1,1,1}][\binom44\binom42\binom41] =41184 $
$3-2-2:[\binom{13}{1,2}][\binom41\binom42^2] =123552$
$3-2-1-1:[\binom{13}{1,1,2}][\binom43\binom42\binom41^2] = 3294720 $
$2-2-2-1: [\binom{13}{3,1}][\binom42^3\binom41]=2471040$
$2-2-1-1-1:[\binom{13}{2,3}[\binom42^2\binom41^3] = 29652480 $
$2-1-1-1-1-1:[\binom{13}{1,5}][\binom42\binom41^5]= 63258624$
Adding up and dividing by $\binom{52}{7}$, we get $\frac{98841600}{133784560}, \approx 0.7388$
A: Your count multiply counts some combinations but not others. For example, you multiply count $AABBCDE$.  Also you divide by $7!$, but you should only divide by $5!$, to reflect the symmetry between the ways to choose the last five cards (that symmetry does not extend to the pairs).  Note, for illustration, that we can count the hands that have a unique pair, that is hands of the form $AABCDE$ as $13\times \binom 42 \times 4^5\times \binom {12}5=63258624$ which is already $.47283$ of the total.
Here's a way to do the count:
The "bad" hands are of the following types:
No duplicates is $4^7\times \binom {13}7=28114944$.
One triple:  Pattern is $AAABCDE$ so $13\times \binom 43 \times 4^4 \times \binom {12}4=6589440$
One quadruple:  Pattern is $AAAABCD$ so $13 \times 4^3 \times \binom {12}3=183040$
Two triples:  Pattern is $AAABBBC$ so $\binom {13}{2}\times \binom 43^2\times \binom {11}1\times \binom 41=54912$
One Quadruple, One triple:  Pattern is $AAAABBB$ so $13\times 12 \times \binom 43=624$
Thus there are $34949260$ bad hands altogether, so $\boxed{133784560-34949260=98841600}$ good hands.
Using these numbers we get the desired ratio:  $$\frac {98841600}{133784560}=\fbox {.7388171}$$
A: Perhaps the following solution using the in-and-out formula is slightly less tedious.
For $1\le i\le13$ let $A_i$ be the set of all $7$-card hands containing an exact pair of rank $i.$
$$|A_i|=\binom42\binom{48}5.$$
$$|A_i\cap A_j|=\binom42\binom42\binom{44}3\text{ for }i\lt j.$$
$$|A_i\cap A_j\cap A_k|=\binom42\binom42\binom42\binom{40}1\text{ for }i\lt j\lt k.$$
By the in-and-out formula,
$$|A_1\cup\cdots\cup A_{13}|=\binom{13}1\binom42\binom{48}5-\binom{13}2\binom42\binom42\binom{44}5+\binom{13}3\binom42\binom42\binom42\binom{40}1$$
$$=133559712-37189152+2471040=\boxed{98841600}.$$
