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I'm reading Advanced Combinatorics by Mitchell T. Keller and William T. Trotter and it's missing an answer book (or, at least, I couldn't find one.) I've been doing the exercises but want to make sure I'm crystal on the content. Here are my questions:

Suppose they do this by selecting a specific type of donut for each employee. (He can select the same type of donut for each person. In how many ways could you do this?)

Since there are 5 employees and 12 donuts to choose from without any specific restrictions, the answer is $12^5$.

How many ways could he select the donuts if he wants to ensure that he chooses a different type of donut for each person?

If he is buying for the employees, there are 5 people and 12 donuts to choose from. Then, the answer is $P(12, 5) = \frac{12!}{7!} = 12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 = 95,040$ ways.

Suppose instead that he wishes to select one donut of each of six different types and place them in the break room. In how many ways can he do this? (The order of the donuts in the box is irrelevant.)

For this last question I'm having difficulty understanding the wording. Is the manager splitting the 12 donuts into 6 categories, therefore there are 2 types of donuts in each category (or partitioning if you will)? And then he is choosing?

Then, if that's the case, $P(2,1)^6 = 2^6 = 64$?

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  • $\begingroup$ Off tangent. I'm inclined to view this a trick question. The number of ways to buy the donuts is different than the number of ways to take the donuts and give them out. If the boss wants to give Martha a maple bar and Jerry a cherry sprinkle then that boss buys one maple bar and one cherry sprinkle. During the buying it doesn't matter who will eventually get what. .... (But I'm pretty sure the intended answer is as you stated.) $\endgroup$ – fleablood Jul 21 '20 at 18:14
  • $\begingroup$ @fleablood So in the case of the buying process, it's just the number of ways to choose six different types of donuts and place them in the break room? $\endgroup$ – dansidorkin Jul 21 '20 at 18:18
  • $\begingroup$ @fleablood Ok, so hypothetically he chooses to hand the donuts out after he bought the donuts. How could I combine the two answers? I.e, if I choose six different types of donuts (given 2 donuts in a partition) how could I then hand them out based on what I have chosen? $\endgroup$ – dansidorkin Jul 21 '20 at 18:19
  • $\begingroup$ Would I multiply the number of ways to choose 6 donuts without replacement for the six people by the number of ways to choose 6 donuts from those size-2 partitions? So the answer would be $6! * 64$? $\endgroup$ – dansidorkin Jul 21 '20 at 18:22
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    $\begingroup$ It seems to me that the answer is simply $\binom{12}6$. Buy $6$ different kinds of donuts and stick them in a box. $\endgroup$ – saulspatz Jul 21 '20 at 18:26
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Forget my smart ass comments. The book is assuming the boss has a list of employees (himself included so there are six employees) one after another. He buys donuts one by one each one specifically for a specific employee.

There are $12$ Types of donuts.

How many ways to select donuts for the employee.

Each employee has a choose of $12$ types of donuts. There are $12^6$ ways to do this. (Don't forget he is buying a donut for himself as well)

How many ways to select donuts for the employess if each employed gets a different type.

$P(12, 6)$ there are $12$ chooses for the first employeed, $11$ for the second....etc.

How many ways to get six different types but the boss doesn't care about the order or who gets what.

"For this last question I'm having difficulty understanding the wording."

I don't see why. This simply a matter of choosing $6$ types of donuts from $12$ different types. THat is ${12 \choose 6} =\frac {12!}{6!6!}$

" Is the manager splitting the 12 donuts into 6 categories, therefore there are 2 types of donuts in each category"

Uh... no. He is doing nothing of the sort. He is buying $6$ (not $12$) donuts and putting them on a table and walking away.

==== post script =====

On rereading question 3 is badly worded "he wishes to select one donut of each of six different types" makes it sound as though the store only had $6$ types not $12$. But still I think it only makes sense as how bought six different types of donuts (one each).

And my smart ass comment is that in buying the donuts order doesn't matter. It's only in passing out the donuts that order matters. But that's me being a smart ass.

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  • $\begingroup$ I think we can all agree question 3 was a badly worded. Just because we have 12 donuts in the donut shop and he wants to select from 6. $\endgroup$ – dansidorkin Jul 21 '20 at 19:07

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