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A cookie baker packages cookies $3$ to a pack. The type of cookies she can choose from include chocolate chip, oatmeal, sugar-coated, sugar-free, peanut butter, and hazelnut. How many different packs of $3$ cookies can she package?

When I approached the problem, I assumed that you had as many of each cookies as you wanted. So the first way to choose the first cookie is $6$, the second is $6$, and the third is $6$ as well. So the total number of possible permutations is $216$. To resolve the order, divide by $3!$ which then yields a total of $36$.

The answer is however $56$ possible packages. I would appreciate that you not only give you provide the correct analysis and solution, but also where I made a mistake.

Ian L
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1 Answers1

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We assume the order of cookies in the package doesn't matter.

There are $\binom{6}{3}$ three-flavour packages. To count the two-flavour packages, the majority cookie can be chosen in $6$ ways, and for each of these ways the minority cookie can be chosen in $5$ ways. Finally, there are $6$ one-flavour packages, for a total of $56$.

Remark: Dividing $216$ by $3!$ is not right. The division by $3!$ is correct for three-flavour packages, but two-flavour ordered packages only come in $3$ orders, not $3!$, and one-flavour packages come in only one order.

André Nicolas
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  • Good solution, although I would've used stars and bars. But if you have to explain that, this answer would've been three times as long. – Arthur Jul 26 '16 at 05:04
  • @Arthur: I considered it, but sum $3$ is "too small" for Stars and Bars, and this particular problem likely comes fairly early in the game, so organized counting feels like the appropriate tool. – André Nicolas Jul 26 '16 at 05:11
  • It's not "too small", you still get $\binom{3+5}{3}$. But yes, it's such small numbers that directly counting is more than enough. – Arthur Jul 26 '16 at 05:29
  • Oh I see. I misread the problem to begin with. Besides, my solution is probably incorrect for what it was intended for. – Ian L Jul 26 '16 at 06:33
  • Would it be equally valid (even though it is numerically) for the second case to say that there is 6 of the first, 6 of the second, and 5 of the third but since there are six ways to arrange each set there is 6 * 5 possible outcomes? – Ian L Jul 28 '16 at 04:04
  • @IanLimarta: I do not understand the logic of $(6)(6)(5)/3!$. If I try to use a similar formula for $8$ types of cookies, I get $(8)(8)(7)/3!$, clearly not right. For ordered two of a kind and one of another, I get $\binom{3}{2}(6)(5)$, and then we divide by $3$. But it is simpler to work directly without order, as I did in the answer. – André Nicolas Jul 28 '16 at 04:26
  • @AndréNicolas I remembered about this problem but then thought about how can the second case be (obviously it is right but I am still confused I suppose). You said that there are 6 ways to choose the first type and then 5 way to choose the second. Isn't this a permutation instead of a combination? Why don't you divide by 2? – Ian L Aug 07 '16 at 01:12
  • If we will have 2 of cookie A and 1 of cookie B, that is different from 1 of cookie A and 2 of B. – André Nicolas Aug 07 '16 at 01:59