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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.

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    $\begingroup$ @BrianMScott last line implies that $0\in\mathbb N$. $\endgroup$ Mar 13 at 21:09
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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 $.

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  • $\begingroup$ thanks for your answer , it is correct but , i hope to find easier way than you did.Because , for large numbers , this solution will be cumbersome process. By the way , i glad to see you again:) +1 $\endgroup$
    – Bulbasaur
    Mar 13 at 21:26
  • $\begingroup$ @Bulbasaur Have you found an easier way? $\endgroup$
    – user
    Apr 17 at 21:42
  • $\begingroup$ I think that an elegant generating function can be found to get rid of cumbersome process and I appreciate your work for me. By the way , i forgot this question. I guess the new answer wont be come. If I can find nice way , i will say you. Thanks for your work.At last , lets forget the past bad experience between us. I appreciate your knowledge and good wills. Have a nice day.. $\endgroup$
    – Bulbasaur
    Apr 18 at 10:16

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