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.

No you are not correct in your final steps. Here is how you could proceed instead: first choose two complete pairs out of $8$ (as you did), for $\binom82=28$ choices. Then choose two pairs from the remaining $6$ for which you will (later) make two incomplete pairs, for $\binom62=15$ choices. Finally from each of those two incomplete pairs, choose one shoe, for $2^2=4$ choices. Every final outcome can be obtained in exactly one way (this is where it differs from your method) and one gets $28\times15\times4=1680$ possibilities.
You found twice that. That is because you found each solution in two different ways, for the $2!=2$ different orders in which you can choose the two unpaired shoes.
