discrete mathematics combinatorics - counting 
I have $52$ cards from $4$ series ($13 \times 4=52 \text{ cards}$). Each series are
  numbered from $1$ to $13$. In how many possible ways you can draw from the
  deck $6$ cards, so that all your six cards has two numbers exactly?

My answer:
First card - $\binom{13}{1} \cdot \binom{8}{4}$ (choosing from the $8$ possibilities, $4$ places I can put my cards. choosing $4$ because I can't have more than $4$ cards of the same number)
Second card - $\binom{13}{1} \cdot \binom{7}{4}$ (same, but $7$)
Total - First card + Second card
Not quite sure about my answer. I am pretty sure I am counting more than there is. Help will be much appreciated.


The number of different  ways  to  obtain $2$ out of $13$ different  numbers is
  \begin{align*}
\binom{13}{2}
\end{align*}

Next we  consider all valid configurations which can be obtained with two numbers.

Since there are always four cards with the same number and we take six cards from the deck there are three types of valid configurations
  
  
*
  
*$(4,2)$:  Four cards with the first number and  two  cards with the second number. There are
  \begin{align*}
\frac{6!}{4!2!}=\binom{6}{2}
\end{align*}
  different configurations of this type.
  
*$(3,3)$: Three cards with the first number and three cards with the second number , giving
  \begin{align*}
\frac{6!}{3!3!}=\binom{6}{3}
\end{align*}
  different configurations.
  
*$(2,4)$:  Two cards with the first number and  four  cards with the second number, giving $$\frac{6!}{2!4!}=\binom{6}{2}$$ different configuration as in the first case.

We conclude:

There are 
  \begin{align*}
\binom{13}{2}\left(2\binom{6}{2}+\binom{6}{3}\right)=3900
\end{align*}
  different configurations to draw six cards with two numbers from the deck.

