Question 1132949
Using notation nCr = n!/((n-r)!r!)<br>

The number of three of kind hands (where three cards match and the other two do not match) is:
 {{{ (13C1)(4C3)(12C2)(4C1)^2 = 13*4*66*16 = 54912 }}}<br>

The factors are:
 (selection of rank)*(#arrangments of three matching cards)*(selection of nonmatching two cards)*(number of arrangements of those two)<br>

The number of full houses (three matching and other two cards match):
  {{{ (13C1)(4C3)(12C1)(4C2) = 13*4*12*6 = 3744 }}} <br>

Thus the total number of hands where three cards match, regardless of the remaining two cards is  54912+3744 = 58656<br>

But, the problem only asks about jacks, which means the above number must be divided by (13C1) since the rank is chosen for you:<br>

58656/13 = 4512<br>


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The number of hands with exactly three jacks is {{{ highlight(4512) }}}

(This also answers the question about how many hands have exactly three queens, as the rank does not matter).  

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Thanks tutor rothauserc!  Yes, that approach is much more direct!  I had poker hands in mind and went the round-about way.   For the student, the first factor in tutor rothauserc's solution is the number arrangments of 3 jacks out of 4, and the 2nd factor is the number of arrangements of the remaining 2 cards out of 48.  What I like about this website is I get to learn too :-)