Say I want to have a eight card hand consisting of three cards that share a rank, 3 cards that share a different rank, and two cards that share a different rank. How many different eight card hands do we have?
$$\binom{13}{1}\binom{4}{3}\binom{12}{1}\binom{4}{3}\binom{11}{1}\binom{4}{2}$$
I seem to be over counting by $2$. I was wondering why this is? And if someone could explain to be where and why I am over counting. Also, are my choices right?
3 Kings -- 3 Queens are counted twice, once when the Kings are the first rank (i.e. number) counted, and once when the Queens are the first rank counted.
The correct approach is to compute the number of distinct ways of choosing the two ranks out of 13 that will each have a triplet :: $\binom{13}{2} = \frac{(13)!}{(11)!(2)!}.$
Other than that, everything else in your answer looks good, including your use of $\binom{4}{2}$ to designate the number of distinct ways of choosing two cards of the same rank (i.e.two cards out of 4).
Note that the reason that this overcounting trap exists in your query is that two of the ranks will have triplets. If instead, you wanted (for example) 1 rank - quadruplet, 1 rank - triplet, and 1 rank pair, then your computation of the ranks as $(13 \times 12 \times 11)$ would be correct.