I have another math tutoring question someone can hopefully help me with?

a, b, c, and d are all non-negative integers. How many total possibilities exist if a+b+c+d=10? I could do it systematically, but that seems like way more work than necessary. Also, I can obviously find the total number of possibilities, but then how would I (easily) eliminate the possibilities which wouldn't add up to 10?

I have another math tutoring question someone can hopefully help me with?
I think this is known as the "number partitioning problem" and I think there is no simple solution to this famous problem. I think you just have to count the ways: Do it systematically:





Here's a better way to count:





Highest Number .... #s used ..... Number of ways


7 .......................... 7 1 1 1 ...............4


6 ...................... 6 2 2 1 ...............4!/2 = 12


5 .................... 5 3 1 1 .............4!/2=12


...................... 5 2 2 1 .......... 4!/2 = 12


4 ................ 4 4 1 1 ........... 4!/(2x2) =6


................ 4 3 2 1 ............ 4! = 24


3 .................. 3 3 3 1 .......... 4


................. 3 3 2 2 ........... 4!/(2x2)=6





Now add up the ways %26amp; that should be the answer


I get 80. Check all my work to see if that's right.
Reply:Brute force:


7 1 1 1 (x 4)


6 2 1 1 (x 12) 16


5 3 1 1 (x 12) 28


5 2 2 1 (x 12) 40


4 4 1 1 (x 6) 46


4 3 2 1 (x 24) 70


4 2 2 2 (x 6) 76


3 3 3 1 (x 4) 80 total possibilities


if a ≠ ≠b ≠ c ≠ d, there are only 24 possibilities.
Reply:You could try solving simpler problems and looking for a pattern. For example if the sum were 0, there's only 1 way.


Sum 1, 4 ways


Sum 2, 10 ways


Sum 3, 20 ways


Sum 4, 35 ways


Sum 5, 56 ways





This is a cubic pattern since the third differences are equal. My TI-84 says it's n^3/6 + n^2 + 11n/6 + 1 where n is the sum.





So 10^3/6 + 10^2 + 11(10)/6 + 1 =


286 ways.
Reply:just do it out in your head its not that hard. Key word: integers.
Reply:Eliminate any possibilities using 9, 8, 7, 6, or 5 because if they were added to integers that were each unique numbers you'd get a number that's larger than ten.





that leaves 4, 3, 2, 1 which equal ten.





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ah, having read your additional detail, I defer to the comments below. Sorry to have added a constraint that wasn't actually in place. Good luck!