Poker Hands and Pairs So I've been to the Wikipedia page for the poker hands and I assume people who know more about this stuff than I do have worked it all out.  What I can't seem to understand is why I can't use the following to calculate the frequency (i.e. total number of times it happens) of getting three of a kind out of a standard deck of cards after being dealt 5 cards as simply:
13 (ranks of cards) * (4 choose 3) * (49 choose 2) 
4 choose 3 being the combination of triples and 49 choose 2 being the combination of all the remaining cards.  I'm aware that this solution would also include four of a kinds and full houses, which the Wikipedia page tallies separately, however the frequency that I get using the above method is: 61,152 vs the Wikipedia response of 59,280 (once full houses and four of a kinds are included). 
Basically I'm just looking to calculate the frequency of having AT LEAST three of the same card.
The problem is further exacerbated if you try the method I was using with pairs giving a 58% (or so) chance of pairs (at least) in a 5 card hand versus the expected 49% (less if you subtracted flushes and straights).
I just want to understand why it doesn't work.  Thanks.
 A: You count the number of hands having at least three of a kind and if several instances of three of a kind occur one of them selected. That way, a hand with four of a kind is counted four times while it should only be counted once.
When doing the same for pairs, the error is even biger because each three of a kind hand gets counted three times, each four of a kind hand gets counted 6 times and each full house hand gets counted three times instead of just once.
A: Take for example hands with at least $3$ Queens. Your method of counting would give $\binom{4}{3}\binom{49}{2}$.  However, this overcounts the $4$ Queen hands. For your $\binom{4}{3}$ counts as one possibility Queens of Spades, Hearts, Diamonds, while your $\binom{49}{2}$ counts, in particular, the hand with the Queen of Clubs and the $2$ of Diamonds. 
But your $\binom{4}{3}$ also allows for the possibility that the three Queens are the Queens of Hearts, Diamonds, and Clubs. Then among the $\binom{49}{2}$  you have counted the Queen of Spades and the $2$ of Diamonds. So each $4$ Queen hand is counted $\binom{4}{3}(48)$ times, and it should only be counted once. 
You can count the at least $3$ of a kind hands as follows:
(i) Exactly $3$: $(13)\binom{4}{3}\binom{48}{2}$ (this will include the full house hands).
(ii) Exactly $4$: $(13)\binom{48}{1}$.
Now add. 
