Need Aid On This Problem Set (Combinations of Month Numbers in 10 Years)

AI Thread Summary
The discussion revolves around a mathematical problem requiring combinations of month numbers (1-12) over a 10-year period to total 49 using exactly eight selections. Participants express confusion over the phrasing, particularly the relevance of the 10-year timeframe, which complicates understanding of the problem. Clarifications suggest that the core issue is to find how many combinations of eight numbers can sum to 49, allowing for repetition. One contributor successfully utilized a SUM calculator to derive the necessary combinations, indicating that the problem was ultimately resolved. Clearer wording in the initial query was emphasized as crucial for obtaining effective assistance.
BBBOB
Messages
5
Reaction score
1
Misplaced Homework Thread -- Moved to the Schoolwork forums
TL;DR Summary: Eight Exact To Get 49

Please advise if I am wording this problem correctly and what are the solution (is there some equation for combinations )/ Answers:SET of numbers 1,2,3,4,5,6,7,8,9,10,11,12 months of the year.

Within the exact time frame of 10 years I MUST choose a number of the month each year that when combined with other months (their number placement in the calendar) over those ten years in total add up to 49.

--------------------------
Constraints:
I must to use exactly eight of the above numbers when added together produce the number 49 over those ten years

Numbers can be used over (repeated) but MUST be eight in total over those ten years.

Questions:

1) How many combinations are there that will equal 49
2) What are those number combinations of eight numbers when added together that equal 49? over those ten years.
---------------------------I'm looking for how many combinations there are

Example:

10,7,5,7,7,4,1,8 ----- these eight when added get 49

10,9,8,7,6,5,3,1 also adds up to 49
 
Physics news on Phys.org
You could write a computer programme to calculate this?
 
  • Like
Likes FactChecker
BBBOB said:
Please advise if I am wording this problem correctly
No you are not: the constraint "Within the exact time frame of 10 years" is not explained.
 
  • Like
Likes FactChecker
PeroK said:
You could write a computer programme to calculate this?
In British English a "computer programme" would be something you watch on television about computers, in the same way as The Sky at Night is an "astronomy programme". When you programme a computer, you write a computer program.
 
I'm not skeptical, I'm sceptical!
 
BBBOB said:
Within the exact time frame of 10 years I MUST choose a number of the month each year that when combined with other months (their number placement in the calendar) over those ten years in total add up to 49.
What does this mean?
BBBOB said:
Example:
10,7,5,7,7,4,1,8 ----- these eight when added get 49
10,9,8,7,6,5,3,1 also adds up to 49
What do these examples have to do with "the exact time frame of 10 years"?
pbuk said:
No you are not: the constraint "Within the exact time frame of 10 years" is not explained.
I agree. As stated, the problem doesn't make sense.
 
pbuk said:
No you are not: the constraint "Within the exact time frame of 10 years" is not explained.
Yes, this is very confusing. How is month 8 of one year different from month 8 of another year? What does the year have to do with anything in the problem?
 
Respectfully disagree with Mark44 and pbuk. The term "combinations" suggests that e.g. 1, 3, 5, 6, 7, 8, 9, 10 will suffice for one partition of 49. The term "10 years" suggests that these numbers, in that order, must be entered in 10 slots or less. That says the finite sequence mentioned can be entered in any of 10C2 = 45 ways. You guys are far above me mathematically; but I'm on another board, where ill-posed problems are par for the course. Thanks for the chance to comment!
 
  • #10
Let's wait and see if @BBBOB comes back to clarify this problem.
 
  • Like
Likes SammyS and PeroK
  • #11
Ben2 said:
Respectfully disagree with Mark44 and pbuk. The term "combinations" suggests that e.g. 1, 3, 5, 6, 7, 8, 9, 10 will suffice for one partition of 49. The term "10 years" suggests that these numbers, in that order, must be entered in 10 slots or less. That says the finite sequence mentioned can be entered in any of 10C2 = 45 ways. You guys are far above me mathematically; but I'm on another board, where ill-posed problems are par for the course. Thanks for the chance to comment!

The 8 in total over 10 years makes no sense though. Why not just 8 years?
 
  • #12
Sorry for the confusion and its mu wording but i was able to use a SUM calculator to get all the combinations that I needed. So this was solved! Thank You Bob
 
  • #13
BBBOB said:
Sorry for the confusion and its mu wording but i was able to use a SUM calculator to get all the combinations that I needed. So this was solved! Thank You Bob
What exactly was the problem you were trying to solve?
 
  • #14
Mark44 said:
What exactly was the problem you were trying to solve?
Trying to see what combinations in only eight digit places add up to 49. Project Im working on. but I have another question which I will post related to such. Thanks! Bob
 
  • #15
BBBOB said:
Trying to see what combinations in only eight digit places add up to 49. Project Im working on. but I have another question which I will post related to such. Thanks! Bob
It would have been helpful and less confusing if you had included this in your original post. The business about "Within the exact time frame of 10 years" was a red herring that confused most of the people replying.

A clearer problem description might be something like this:

Using the numbers 1, 2, 3, ..., 12, how many combinations of exactly eight of these numbers add up to 49? A combination can include repeats of these numbers.​

If you want to get good answers, it helps to ask a good questions.
 
  • #16
Point taken, you are so right. In math how you word it is crucial! Thank you again!
 
  • #17
This seems like your standard "Stars and Bars " problem,
https://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)
if I understood it correctly.
##x_1 +x_2 +....+x_k= n; x_i \geq 0 ##* The answer, with ## x_i >0 ## for the count (**)
is ## C_{n-1,k-1} ## ("## n-1 ##choose ##k-1##"), though it won't exclude ##x_i >12##
Here, ##C_{49-1, 4-1}=C_{48,3}=17,296 ##

* There's a version that requires ## x_i >0 ##t
** Or the countess, to be inclusive ;)
 
Last edited:
  • #18
Very good thank you all for your input!
 

Similar threads

Replies
7
Views
3K
Replies
2
Views
3K
2
Replies
93
Views
14K
3
Replies
104
Views
16K
Replies
1
Views
2K
Back
Top