[Raymundo]

Your analysis is quite correct when you are only considering three cards total. But here the situation is a little trickier because it's the best three out of five. I've elaborated below, but I'll warn you that it might not be worth the effort or the time it takes to read it.

For trips, I counted any normal five card poker hand that has trips in it, specifically all quads, full houses, and trips. That added up to
Quads = 624 possibilities
Fulls = 3,744 possiblities
Trips = 54,912 possibilities
TOTAL = 59,280 five card hands that have trips in them.

The sf's were trickier. An sf is completely described by its lowest card. There are 12 sf's from A to Q.
-Sf's beginning with A thru J can get their other two cards from any 48 of the remaining 49 cards (the 49th card would be the next highest card in the sf and so you can't count it twice.) There are four different suits to choose from so
12 sf's x 4 suits x 48C2 blanks = 54,144 sf's
-Sf's beginning with the Q can get their other two cards from all 49 of the remaining cards.
1 sf x 4 suits x 49C2 blanks =4,704
Thus, TOTAL five card hands containing a three card sf is 58,848.

So it appears that there are slightly less sf's than trips. But here's the problem: of the 59,280 hands including trips, 432 of them have both trips AND sf's. Since sf's are being counted higher, it turns out that we can't count them as trips hands, only as sf hands. Deducting 432 then from 59,280 leaves us with 58,848 WHICH IS EXACTLY HOW MANY SF HANDS THERE ARE!