# How many solutions for equation, with Inequation (Combinatorics)

Encountered this question, with a solution, that I didn't understand...

How many solutions are there for the equation: $$x_1+x_2+x_3+x_4+x_5+x_6=24$$ Given that $$x_1+x_2+x_3>x_4+x_5+x_6$$

There is a hint:

" A solution for the equation should be exactly one of these: $$x_1+x_2+x_3>x_4+x_5+x_6$$ $$x_1+x_2+x_3<x_4+x_5+x_6$$ $$x_1+x_2+x_3=x_4+x_5+x_6$$ "

The solution for the question says that if the requested number is $x$ , then because of symmetry: $2x + (D(3,12))^2 = D(6,24)$

• Note: I don't know if it's a common symbol, but in our course $D(n,k) = \binom{n-1+k}{k}$

I understand that $D(6,24)$ is the number of solutions to the original equation, without the additional ineqaulity info.

I understand that $D(3,12)$ is the number of solutions for $x_1+x_2+x_3=12$

But how do I continue? I don't understand the other parts of the solution...

Addition is commutative and associative so there's nothing in the expression that distinguishes any $x_i$ from the other. So any fact about them should hold true under a permutation of the indices. In particular,

The number of solutions for which

$$x_1 + x_2 + x_3 > x_4 + x_5 + x_6$$

should equal the number of solutions for which

$$x_4 + x_5 + x_6 > x_1 + x_2 + x_3$$ (which can be re-written as: $x_1 + x_2 + x_3 < x_4 + x_5 + x_6$)

The reason why you need $D(3,12)^2$ is because when $x_1 + x_2 + x_3 = x_4 + x_5 + x_6$, we know that $x_1 + x_2 + x_3 = 12$ and $x_4 + x_5 + x_6 = 12$. Because we know how many solutions there are to each of these, we can multiply the number to get how many possible combinations there are for all $x_1, ..., x_6$ in this case.

• and why $(D(3,12))^2$? – Zvika Sep 13 '17 at 15:32
• I'll edit the answer – muzzlator Sep 13 '17 at 15:32
• great. And why can I be sure that there is such symmetry? shouldn't I prove it? – Zvika Sep 13 '17 at 15:33