Not sure if I understand the question:

If $x$, $y$ and $z$ are three integers then since I already have $(1,2,9)$ is $(1,9,2)$ not allowed?

Does the question implicitly limit the largest integer to $9$?

I tried the following but since the answer is $12$ I'm not certain if my approach is right:

1 Series:-
$$(1,9,2),\ (1,2,9),\ (1,8,3),\ (1,3,8),\ (1,7,4),\ (1,4,7),\ (1,6,5),\ (1,5,6)$$

2 Series:- $$(2,4,6),\ (2,6,4),\ (2,3,7),\ (2,7,3),\ (2,9,1),\ (2,1,9)$$

3 series:- $$(3,5,4),\ (3,4,5),\ (3,7,2),\ (3,2,7),\ (3,1,8)\ (3,8,1)$$

4 series:- $$(4,3,5),\ (4,5,3),\ (4,2,6),\ (4,6,2),\ (4,1,7),\ (4,7,1)$$

5 series:- $$(5,3,4),\ (5,4,3),\ (5,1,6),\ (5,6,1)$$

6 series:- $$(6,4,2),\ (6,2,4),\ (6,5,1),\ (6,1,5)$$

7 series:- $$(7,3,2),\ (7,2,3),\ (7,4,1),\ (7,1,4)$$

8 series:- $$(8,3,1),\ (8,1,3)$$

9 series:- $$(9,2,1),\ (9,1,2)$$

If I consider only the unique set of numbers then I seem to be getting only $7$:

$$(1,9,2),\ (1,8,3),\ (1,7,4),\ (1,6,5),\ (2,4,6),\ (2,7,3),\ (3,5,4)$$

But the answer is $12$

Is "Trial & Error" the only way to figure the answer to this problem or is there a "Method" to this problem?

I have to explain this to my 3rd grader son and any help would be highly appreciated.

Question: How many different sets of three positive integers have a sum of $12$?

Answer: $12$

  • $\begingroup$ Do the three integers have to be different? So 1+1+10 is not valid? $\endgroup$ – turkeyhundt Sep 22 '16 at 16:42
  • $\begingroup$ It looks like $1+1+10$ is counted as a set of three integers adding upto 12. Poorly formulated question. $\endgroup$ – P Vanchinathan Sep 22 '16 at 16:44
  • $\begingroup$ Also, if you want to research this, one phrase you can look into is, "indistinguishable balls into indistinguishable boxes" which is equivalent to partitioning a number $m$ into $n$ parts. $\endgroup$ – turkeyhundt Sep 22 '16 at 16:48
  • $\begingroup$ If you are allowed repeated numbers and order matters, this becomes a standard question of Stars-and-Bars. If order doesn't matter and either you are or aren't allowed to repeat numbers, this becomes a restricted partition problem however I would not expect anyone to see either of these in the third grade. $\endgroup$ – JMoravitz Sep 22 '16 at 16:49
  • 1
    $\begingroup$ speaking of sets, (1,2,9) and (1,9,2) are coincident. $\endgroup$ – G Cab Sep 22 '16 at 17:00

The number of partitions of $12$ of size $3$ is $12$: http://www.wolframalpha.com/input/?i=partitions+of+12++of+size+3

That's the number of ways to choose three positive integers that add up to $12$, allowing the terms of the sum to be repeated but not distinguishing between permutations of a list of integers. For example, $4+4+4 = 12$ is a partition of $12$ of size $3$, and so is $8+2+2= 12$.

This may be the interpretation that was intended. If the question has copied the words of the original problem exactly, it may be that the problem was poorly worded. The question seems really to be asking for how many distinct multisets of positive integers exist that each have a sum of $12$.

  • $\begingroup$ Question is an exact copy from the original:- amazon.com/Challenge-Problems-Elementary-Middle-schools/dp/… $\endgroup$ – Gopinath Rajee Sep 22 '16 at 17:51
  • $\begingroup$ I've noticed quality problems in books at about this level of education. This is not nearly the worst example in my opinion. $\endgroup$ – David K Sep 22 '16 at 19:15
  • $\begingroup$ David, Would you be able to give me a list of such books that you might be aware off? It would be of great help for me to teach my 3rd grader son. $\endgroup$ – Gopinath Rajee Sep 23 '16 at 15:59
  • $\begingroup$ Mostly I notice these things when someone asks MSE about a problem in one of their textbooks, on an on-line assignment, or in a test, and it is obvious that the problem was flawed. These are all examples of bad questions, and in most cases I didn't read much of the rest of the textbook and can't comment on its overall quality. I'm not sure what really can be taught to a 3rd grader from any of this; "never trust your textbook" is a lesson that might occasionally pay off, but I'm afraid it would more often be a hindrance to learning. $\endgroup$ – David K Sep 23 '16 at 16:30

The elements of a set must be distinct, so there are 7 sets:(1,2,9),(1,3,8),(1,4,7),(1,5,6),(2,3,7),(2,4,6),(3,4,5)


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