The number of ways of selecting 6 shoes from 8 pairs of shoes so that exactly 2 pairs of shoes are formed The number of ways of selecting  6 shoes from 8 pairs of shoes so that exactly 2 pairs  of shoes are formed?
My try:

Let us first choose $2$ pairs from $8$ pairs. It can be done in $_{8}C_2$ ways.
Suppose I have chosen pair $(1,2)$ and $(3,4)$. We have to chose other 2 shoes such that one is not a pair of another. 
For example if I choose 5 I shouldn't choose it's pair ie 6.  It can be chosen in $_{12}C_1 \cdot _{10}C_1$ ways. So total no. of ways = $_8C_2\cdot _{12}C_1\cdot_{10}C_1=3360$ ways. 
Am I correct. Please clarify.

As you said you can choose $2$ pairs out of $8$ pairs in ${8\choose2}=28$ ways. When these $2$ pairs have been chosen there are $6$ complete pairs left, and you have to choose another $2$ shoes out of these such that no additional pair is formed. This is possible in ${12\choose2}-6=60$ ways. The total number of admisible choices therefore is $28\cdot60=1680$.
