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?