Permutation (or Combination?) Word Problem I'm continuing to work through permutations and combinations and my book has the following question:

A box contains 9 red, 4 blue, and 6 yellow chips. In how many ways can
  6 chips be chosen if: 
  a) all 6 chips are red 
  b) all 6 chips are yellow 
  c) 2 chips are blue

I had no problem answering a) and b); however c) is stumping me. The answer, according to the text, is 8,190 however I can't figure out how they're arriving at that number.
I thought it would be:
$$(19\cdot 18\cdot 17\cdot 16/4!)/2! = \frac{19\cdot 18\cdot 17\cdot 16}{4!\cdot 2!}$$
but that returns $1938$, not $8,190$.
At this point in the text we've only covered the Permutation and Combination formulas: nPr and nCr.
Help?
 A: The number of ways of choosing 2 blue out of 4 is $\binom{4}{2}.$  After that you have 4 chips remaining which can be chosen from either  9 red or 6 yellow or both. So you are choosing 4 chips from $9 + 6 =15$ chips.
Putting all together, you get $$\binom{4}{2}\binom{15}{4}.$$
A: You can solve those kind of questions using generating functions:
- For the choice of red chips: $1+x+...+x^9=\frac{1-x^{10}}{1-x}$ (which means you can choose no red chips or 1 or 2 and so on up to 9).
- For the blue chips: $x^2$ since you have to choose $2$ blue chips.
- For the yellow chips: $1+x+...+x^6=\frac{1-x^7}{1-x}$ (once again, you can choose up to 6 yellow chips).
Now, the generating function to your problem would be $f(x)=\frac{1-x^{10}}{1-x}x^2\frac{1-x^7}{1-x}$. What you need is the coefficient of $x^6$, which is the number of ways to choose exactly $6$ chips under the above restrictions.
Hope that helps.
A: You want 2 blue chips out of 6 and 4 non-blue chips out of 15.
So $_6C_2\cdot _{13}C_4 = 15\cdot \frac{15\cdot14\cdot13\cdot12}{4\cdot3\cdot2\cdot1} = 15\cdot (15\cdot7\cdot13)= 20475$.
