# How many solutions are there in the following equation over the natural numbers if $x_1+x_2+x_3>x_4+x_5+x_6+x_7$

How many solutions are there in the following equation over the natural numbers such that $$x_1+x_2+x_3+x_4+x_5+x_6+x_7=30$$ if $$x_1+x_2+x_3>x_4+x_5+x_6+x_7$$ ?

I made up a combinatorics question in my mind , but i stuck in it. What i tried is :

Firstly , i wanted to use symmetry property but , it has odd number of variables . Hence ,i could not do anything.

Secondly , i thought about whether generating functions can be used or not , but i stuck in writing the desired generating formula for it.

So , i hope to find nice approaches to my question.

$$NOTE=$$Natural numbers start with zero.

• @BrianMScott last line implies that $0\in\mathbb N$. Mar 13, 2021 at 21:09

Using stars and bars you can write the number of solution as $$\sum_{i=0}^{14}\binom{i+3}3\binom {30-i+2}2,$$ where the first factor counts the number of solutions to the equation $$x_4+x_5+x_6+x_7=i$$ and the second one those of $$x_1+x_2+x_3=30-i$$.