Farina, in order to explain the concept behind the multiplication, I will use combinatorics (I hope that this approach could help you):

1) How many pairs can we construct (total outcomes)?
\(\frac{40!}{2!(40-2)!} = 780\)
2) how many pairs of left-handed and right-handed can we do?
\(12*28 = 336\)
why?: For each left-handed student we can find 28 students in order to form a couple (and vice versa), therefore, the quantity of total couples of left-handed and right-handed is given by 336.
3) The probability is given by \(\frac{Possible outcomes}{Total outcomes} = \frac{336}{780}=\frac{28}{65}\)

EDIT: The need for multiplying by 2, is because, otherwise, you are considering only one pair, let's say {left-handed,right-handed}, but you have to consider {right-handed, Left-handed} also.

The possibilities of choosing a right-handed and the left-handed are given by:
\(\frac{28}{40}\frac{12}{39}\)
The possibilities of choosing a left-handed and a right-handed are given by:
\(\frac{12}{40}\frac{28}{39}\)

But, those fractions are equal:
\(\frac{12}{40}\frac{28}{39} = \frac{28}{40}\frac{12}{39}\)

The final probability is equal to the sum of both:
\(\frac{12}{40}\frac{28}{39} + \frac{28}{40}\frac{12}{39} = 2*\frac{28}{40}\frac{12}{39} = 2*\frac{14}{65}\)

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