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.

When unsure, try with a smaller instance of the problem.
Let's count the ways of choosing 2 shoes from 2 pairs so that they are not in the same pair. Your method would say start with 0 full pairs out of 2 and then multiply by $_4C_1$ and $_2C_1$:
$$\binom{2}{0} \binom{4}{1} \binom{2}{1} = 8$$
But we can count directly: out of $ABab$, the choices fulfilling our conditions are $Aa$, $Ab$, $Ba$, $Bb$. That's not 8 so the formula is not correct.
At the same time it shows what should have been done differently and why: divide your result by 2 to reflect that the last 2 shoes that must not be from the same pair can be picked in any order ($aA = Aa$).
