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My professor gave us this problem. May you give me your thoughts about my answer and the problem?

Thank you so much!

Here is the problem:

If object A can be chosen with repetition either not at all or twice, object B either once or three times, and object C not at all or once, how many selections of thirteen objects can be made?

I want to explain my reasoning about the generating functions that I selected.

Generating Function for object A:

As object A can be selected with repetition either not at all or twice,

without repetition would be $(1+x^2)$ ($1$ for zero A's, $x^2$ for two A's).

I am assuming that the second time that we choose object A, we can have zero or $4$ A's, then we have $(1+x^2+x^4)$, the third time we have $(1+x^2+x^4+x^6)$ and the maximum quantity of A's would be 12 because, we will have at least one object B.

Thus the generation function for object A would be:

enter image description here

Generating function for object B:

As object B can be selected either once or three times, I will be using x for once and $x^3$ for three times. Then the generating function for object B is:

$(x + x^3)$

Generating function for object C

As object C can be selected not at all or once, I am using

$(1+x)$

Finally, I multiply all of them, using Wolfram Alpha:

Wolfram Alpha calculations

My answer is the coefficient of x^13 that is 2.


I want to try a

New approach:

The possible values for A are 0,2,4,6,8,10 and 12.

The possible values for B are 1, 3

The possible values for C are 0, 1

The combinations for B and C are only four:

0 C's + 1 B's (Subtotal 1 object, the only possible values for A is 12)

0 C's + 3 B's (Subtotal 3 objects, the only possible value for A is 10)

1 C's + 1 B's (Subtotal 2 objects, there are no possible values for A to complete 13)

1C's + 3 B's (Subtotal 4 objects, there are no possible values for A to complete 13)

This approach also confirms that there are only 2 options.

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    $\begingroup$ What other objects are in there. If AB and C are the only objects this can't be done because you can't pick any object more than three times. If there are 7 other objects there is only 1 way. If there are 1000 other objects there are many ways. $\endgroup$ – fleablood Jun 5 '17 at 6:11
  • $\begingroup$ @fleablood Thank you for your help! Sorry for the delay.... I was sleeping... About your question it seems that is possible because the problem allows repetition. It seems that we start with object B, then we choose randomly objects A, B or C, and then we repeat until we choose the 13 objects. At least, that is how I understand the problem $\endgroup$ – Beginner Jun 5 '17 at 12:18
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    $\begingroup$ I don't understand your answer to fleablood's comment. Please give an explicit example of $13$ choices that satisfy your restrictions (and maybe an example that does not satisfy them). E.g., does BAACCBBCAACBB satisfy the restriction? Why or why not? $\endgroup$ – Barry Cipra Jun 5 '17 at 12:27
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    $\begingroup$ @Beginner, I think you're going to have to ask your professor for clarification. I for one am utterly baffled as to what is meant. $\endgroup$ – Barry Cipra Jun 5 '17 at 13:21
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    $\begingroup$ If the answer is 2 then you should by able to list precisely which 2 there are. Which 2 are they? I do not understand generating functions. And I find the question as written utterly incomprehensible so I can't tell you what I think. Maybe if you listed the two examples that work and explain why they work I might understand this better. $\endgroup$ – fleablood Jun 5 '17 at 16:49
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You have a good explanation of the solution to the problem as you understand it. As others have said, it is difficult to understand if this is what the problem is asking for. I think your reading is reasonable.

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  • $\begingroup$ Thank you so much for your help! I am so sorry. I edited my question while your where publishing your answer. May you check if I should go back to my previous version of my answer or if my new answer is better? $\endgroup$ – Beginner Jun 5 '17 at 16:42
  • $\begingroup$ I think your new reading is also reasonable. Again, you have the correct to the answer to the problem you are working on. Whether it is the one your professor intends, only s/he can say. $\endgroup$ – Ross Millikan Jun 5 '17 at 17:21
  • $\begingroup$ Thank you so much for all your time and ideas. $\endgroup$ – Beginner Jun 5 '17 at 17:25
  • $\begingroup$ My professor told me that my LAST interpretation of the problem, that I published is the right one and that my answer is correct. Thank you for all your advices. $\endgroup$ – Beginner Jun 5 '17 at 23:45
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"If object A can be chosen with repetition either not at all or twice, object B either once or three times, and object C not at all or once, how many selections of thirteen objects can be made?"

Depending on interpretation either $0$ or $0$ or $1$.

If "choosing" means total number of times one can chose A at most 2 times, and B at most 3 times and C at most 1 time. That means at least 7 choices must be something else.

If "choosings" means number of times consecutively then one is never allowed to pick $0$ B's so one must always pick $B$ and can not ever not pick $B$. So one must start by picking a $B$. Then either one can pick something else (one $B$) or pick two more $B$ (three $B$). If one picks something else, that is zero $B$ which is impossible. So one must pick two more $B$s. One has now picked $3$ $B$. Now one may pick another $B$ or something else. If one picks another $B$ that is four $B$s and not allowed. If one picks something else that is $0$ $B$s and not allowed. So this is impossible.

If you mean that when you pick one object you must pick it an allowable number of times, but you are allowed to immediately pick it again one of the allowable numbers of times. Then the only acceptable option is $BBBBBBBBBBBBB$. You are never allowed to pick $B$ zero times so you are never allowed not to pick $B$.

I can not see any other interpretation that would be consistent.

Your interpretation that anything except those with no $B$ are allowed would yield $3^{13} - 2^{13}= 1,586,131$ which is more than $49$. If you mean the when you chose $A$ you must chose consecutively and even number of times and if you choose $B$ you must chose it consecutively a number of times that must be a sum of $1$s and $3$ (which means it can by anything) then this is inconsistant with what "not at all" means. This would be $\sum\limits_{n=0}^6 {13 -n \choose 13 - 2n}*2^{13-2n}$.

Frankly, I find this question utterly incomprehensible.

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  • $\begingroup$ Thank you for all the time that you have invested helping me! I appreciate it a lot! $\endgroup$ – Beginner Jun 5 '17 at 16:49
  • $\begingroup$ My professor told me that my LAST interpretation of the problem, that I published is the right one and that my answer is correct. Thank you for all your advices. $\endgroup$ – Beginner Jun 5 '17 at 23:45

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