# 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?

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## 3 Answers

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 $$\binom{4}{2}\binom{15}{4}.$$

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Thanks so much for the textual explanation (not to mention the calculation)! Really helpful. –  Nick Jul 21 '11 at 16:44
No problem. glad to help..:) –  Nana Jul 21 '11 at 16:48
Sorry, one more question on this. After choosing 2 blue out of 4, why would the the 4 chips remaining not be chosen from 9 red, 6 yellow and/OR 2 blue? In other words after you choose the 2 blue chips aren't there still 2 other blue chips remaining that could be chosen to fill the remaining 4 places? –  Nick Jul 23 '11 at 15:31
Never mind. Got it. The problem states only 2 chips are blue, so we don't include them. If the problem said "at least 2 chips are blue" then we'd include the remaining 2 blue chips in the second combination: 17C4, as opposed to 15C4. –  Nick Jul 23 '11 at 15:41
Im glad you got it. If it however said at least 2 blue chips, then we could have 2 blue, 3 blue or 4 blue chips.So that will be (4C2*15C4)+(4C3*15C3)+(4C4*15C2) –  Nana Jul 23 '11 at 16:09

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.

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You want 2 blue chips out of 6 and 4 non-blue chips out of 15.

So $_6C_2\cdot _{13}C_4 = 15\cdot \frac{15\cdot14\cdot13\cdot12}{4\cdot3\cdot2\cdot1} = 15\cdot (15\cdot7\cdot13)= 20475$.

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