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

Question

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?

There's no limitation on these numbers being unique. They can all be the same number. Thus, 7+1+1+1=10 works, and so on. I started to list them out, but there were over 50 possibilities before I stopped. This leads me to believe there's a statistical method that will do this, but I'm blanking on how.

Answer 1

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 & that should be the answer
I get 80. Check all my work to see if that's right.

Answer 2

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.

Answer 3

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.

Answer 4

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!

Answer 5

just do it out in your head its not that hard. Key word: integers.