Counting/Probability of Picking Chairs (Answer validation) 
In a room, there are 16 wooden chairs and 10 plastic chairs. Except for
the color, the wooden chairs are identical and the same holds for the
plastic chairs. Of the wooden chairs, 5 are red, 5 are blue and 6 are
green. Of the plastic chairs, 4 are red, 2 are blue and 4 are green.

Question 1:

In how many different ways can 9 chairs be chosen from the total
number of 26 chairs in the room such that there are 3 of each color?

Question 2:

What is the probability that only one of the 9 chosen chairs is
wooden?

My try for question 1:
$${9 \choose 3}{7 \choose 3}{10 \choose 3} = 352,800$$
My try for question 2:
Let A be the event of picking 1 wooden and 8 plastic chairs.
Let B be the event of picking 9 chairs with exactly 3 colors each.
$$\Pr(A \cap B) = \dfrac{{16 \choose 1}{10 \choose 8}}{{26 \choose 9}}$$
$$\Pr(B) = \dfrac{352,800}{{26 \choose 9}}$$
$$\Pr(A|B) = \dfrac{{16 \choose 1}{10 \choose 8}}{352,800} = \dfrac{1}{490}$$
The probability for question 2 seems too small. Can someone point out my mistake?
 A: As Karl pointed out in the comments, we have to consider the color of the wooden chair when calculating $\Pr(A \cap B)$.  There are three possibilities:

*

*one red wooden chair, two red plastic chairs, three blue plastic chairs, and three green plastic chairs

*three red plastic chairs, one blue wooden chair, two blue plastic chairs, and three green plastic chairs

*three red plastic chairs, three blue plastic chairs, one green wood chair, and two green plastic chairs

Since there are only two blue chairs, the first and third cases cannot occur.  Therefore, the number of ways exactly one wooden chair can be selected if three chairs of each color are selected is
$$|A \cap B| = \dbinom{4}{3}\dbinom{5}{1}\binom{2}{2}\dbinom{4}{3}$$
You correctly found that the number of ways of selecting three chairs of each color is
$$|B| = \binom{9}{3}\binom{7}{3}\binom{10}{3}$$
and that the number of elements in the sample space is
$$|S| = \binom{26}{9}$$
Hence, the probability that exactly one of the nine chairs selected is wooden given that three chairs of each color are selected is
\begin{align*}
\Pr(A \mid B) & = \frac{\Pr(A \cap B)}{\Pr(B)}\\ 
              & = \frac{\frac{|A \cap B|}{|S|}}{\frac{|B|}{|S|}}\\ 
              & = \frac{|A \cap B|}{|B|}\\
              & = \frac{\dbinom{4}{3}\dbinom{5}{1}\dbinom{2}{2}\dbinom{4}{3}}{\dbinom{9}{3}\dbinom{7}{3}\dbinom{10}{3}}
\end{align*}
