A deck of 40 cards, 4 suits from 1 to 10, 2 cards extracted, probability the sum is 9? You have a deck of $40$ cards, $4$ suits from $1$ to $10$, pick randomly $2$ cards, no reimmission, what  is the probability you get $9$ as a sum?
How can I solve using the basic principle of counting?
The favorable outcome are $(8+1)*4$, $(7+2)*4$, $(6+3)*4$ and $(5+4)*4$, so $64$ possible combination?
But what about the all the possible outcome? Are $40*39$?
From a MC simulation the results seems $~0.08$, but I can't do analytically.  
EDIT:
The numerator is wrong, a combination can be also 1 and then 8, 2+7 and so on, the combinations are, actually, doubled, 128.
$$ \frac{128}{40\cdot39} = .0821 $$ 
Thank you all for the answer.
 A: What is the probability that we select $1$ and $8$? There are $4$ $1$'s and $4$ $8$'s and we select $2$ cards from the $40$ so we have
$$\frac{{4\choose{1}} \cdot {4\choose{1}}}{40\choose{2}}$$
This is the same as the probability of selecting a $2$ and a $7$, a $3$ and a $6$, and a $4$ and $5$. 
So the probability that the $2$ cards sum to $9$ is
$$\frac{4 \cdot {4\choose{1}} \cdot {4\choose{1}}}{40\choose{2}}\approx.082$$
Another way to approach this is to consider the probability of selecting a $8$ and a $1$. We can select $8$ and then $1$ or $1$ then $8$ giving a probability of 
$$2\cdot \frac{4}{40} \cdot \frac{4}{39}$$
Then accounting for $1$ and $8$, $2$ and $7$, $3$ and $6$, and $4$ and $5$, we get
$$4 \cdot 2\cdot \frac{4}{40} \cdot \frac{4}{39} \approx .082$$
A: Every card in the deck, less than $9$, has a $9's$ complement that is distinct from that card.  So:
Probability that the first card drawn is in range $1$ to $8$ inclusive:$$P_{1 \text{ to } 8}=\frac{32}{40} = 0.8$$ 
Probability that, given that the first card is in range $0$ to $8$, the second card drawn is the $9's$ complement of the first card:$$P_{Comp}=\frac{4}{39}$$
So the probability of the desired sum is just:$$P_9=\frac{32}{40} \times \frac{4}{39}=\frac{16}{195}$$
