Criticize my math when I attempt to find the coefficient of $x^2y^6$ in the expansion of $(x+2y^2)^5$ So I look around this site and my textbook (Richmond&Richmond, discrete math) and I know I'm in the right direction but I'm also sure I am doing it wrong.
Original Question: find the coefficients of $x^2y^6$ in the expansion of 
                                      $$(x + 2y^2)^5$$
So far I have:
       $$(x+2 y^2)^5 \implies x^2+2^3×y^6  \implies (5!)/(2!×6!)×2^3 = 2/3$$
But that definitely does not seem right. 
I'm also avoiding learning combinations by drawing out Pascal's Triangle since on a time sensitive exam its not practical.
Comments, suggestions, snide remarks; anything to help is appreciated ahead of time. 

There are two aspects in your derivation which should be corrected.

When selecting the terms with $x^2$ and $\left(2y^2\right)^3$ you have to multiply them instead of adding them.
  \begin{align*}
  (x+2 y^2)^5  \implies   x^2\left(2y^2\right)^3=2^3x^2y^6
  \end{align*}
When identifying the corresponding binomial coefficient, note that it is
  \begin{align*}
  (5!)/(2!3!)=\binom{5}{2}=\binom{5}{3}=10
  \end{align*}
You derivation could therefore be
\begin{align*}
(x+2 y^2)^5  \implies   2^3x^2y^6 \implies \frac{5!}{2!3!}\cdot 2^3=\frac{120}{2\cdot6}\cdot8=80
  \end{align*}

Another variation:

In case you already know the binomial theorem you can use
  \begin{align*}
  \left(x+2y^2\right)^5=\sum_{k=0}^5\binom{5}{k}x^k\left(2y^2\right)^{5-k}\tag{1}
  \end{align*}
In order to find the coefficient of $x^2$ we see there is only the term with $k=2$ a candidate. The summand with $k=2$ gives
  \begin{align*}
  \binom{5}{2}x^2\left(2y^2\right)^{5-2}=2^3\binom{5}{2}x^2y^6=80x^2y^6
  \end{align*}
  and we conclude the coefficient of $x^2y^6$   is $80$.

