\(\displaystyle \dfrac{\dbinom{G+3}{2}+\dbinom{G}{2}}{\dbinom{2G+3}{2}}=\dfrac{27}{55}\)

I'm also confused by this question. Please can you explain it further?

The problem is about picking two pens of the same colour. Two blues or two greens. But it is given that "There are three more blue pens than green pens in the box" so \(\displaystyle B=G+3\) Thus there are \(\displaystyle \dbinom{G+3}{2}+\dbinom{G}{2}\) ways to pick two pens of the same colour, either two blue or two green. That also means we have a total of \(\displaystyle 2G+3\) pens in the box.

Lets look at the probability: \(\displaystyle \dfrac{\dbinom{G+3}{2}+\dbinom{G}{2}}{\dbinom{2G+3}{2}}=\dfrac{27}{55}\)

\(\displaystyle \dfrac{\dbinom{G+3}{2}+\dbinom{G}{2}}{\dbinom{2G+3}{2}}=\dfrac{\dfrac{(G+3)(G+2)}{2}+\dfrac{G(G-1)}{2}}{\dfrac{(2G+3)(2G+2)}{2}}=\dfrac{27}{55}\)

Simple algebra should let one solve for \(\displaystyle G\) But remember that \(\displaystyle G+B>12\).